#math/linear-algebra # From linear maps to row reduction This note bridges the basis-independent language of linear maps with the computational language of matrices and systems. These are two descriptions of the same mathematics, not competing theories. The route is: start with the LADR viewpoint, $T:V\to W$; choose bases to obtain a matrix; then use the LADW viewpoint, row operations on $Ax=b$, to investigate the original map. All scalars are real. Work through the checkpoints before opening their solutions. The example deliberately stays simple so that the distinction between vectors, coordinates, and maps remains visible. ## 1. What seems to be getting mixed together Based on the questions so far, the gap seems to involve these distinctions: 1. The range describes which outputs are reachable. A solution set describes which inputs reach one specified output. These are different sets in different spaces. 2. Dimension counts independent directions, not the number of vectors. A plane contains infinitely many vectors; a line does too. 3. A noninjective linear map has a nontrivial nullspace, not an empty nullspace. There are nonzero inputs whose output is zero. 4. A vector and its coordinate list are not intrinsically the same object. Using standard bases usually hides the distinction. 5. Row-equivalent matrices need not give the same outputs in fixed coordinates. They give equivalent systems when the target coordinates undergo the same row operations. The earlier explanation moved too quickly between “a different map in fixed coordinates” and “the same map in different coordinates.” We will separate those interpretations before putting them together. ## 2. Start with the map, before introducing a matrix Write $ T:V\longrightarrow W. $ - $V$: the domain, containing inputs. - $W$: the codomain, containing possible targets. Not every target must be reachable. - $T$: the linear map, assigning an output to every input. - $v\in V$: one input vector. - $Tv\in W$: its output vector. The notation $Tv$ means $T(v)$. There are two different questions: | Question | What is fixed? | What are we finding? | | --- | --- | --- | | Compute $Tv$. | An input $v$. | Its one output. | | Solve $Tv=w$. | A target $w\in W$. | All inputs that produce that target. | A function always gives one output per input. That does not mean there is one input per output. ### How coordinates enter Choose an ordered basis $\mathcal B$ of $V$ and an ordered basis $\mathcal C$ of $W$. - $[v]_{\mathcal B}$: the column of coefficients describing $v$ in the input basis. - $[Tv]_{\mathcal C}$: the column describing its output in the output basis. - $A$: the matrix of $T$ relative to these two bases. The defining relationship is $ \underbrace{A}_{\text{matrix of }T} \underbrace{[v]_{\mathcal B}}_{\text{input coordinates}} = \underbrace{[Tv]_{\mathcal C}}_{\text{output coordinates}}. $ For a target $w\in W$, put $x=[v]_{\mathcal B}$ and $b=[w]_{\mathcal C}$. Then $ \underbrace{Tv=w}_{\text{equation between vectors}} \quad\Longleftrightarrow\quad \underbrace{Ax=b}_{\text{equation between their coordinate columns}}. $ When the spaces are $\mathbb R^3$ and the bases are standard, a vector and its coordinate column have the same entries. We usually identify them. That convenience becomes confusing when a basis changes. ## 3. One running example Let $V$ and $W$ be separate copies of $\mathbb R^3$. - $\mathcal B=(e_1,e_2,e_3)$: the standard input basis in $V$. - $\mathcal C=(f_1,f_2,f_3)$: the standard output basis in $W$. - Subscripts $1,2,3$ label basis vectors, not separate maps. Although the two spaces have the same dimension, we use different basis letters to keep input directions separate from output directions. Notation rule for this example: write actual input vectors using $e_i$, actual output vectors using $f_i$, and coordinate columns using $[\,\cdot\,]_{\mathcal B}$ or $[\,\cdot\,]_{\mathcal C}$. The numbers alone do not say which space or basis is intended. Define $T$ by its action on the input basis: $ T(e_1)=f_1,\qquad T(e_2)=f_2+f_3,\qquad T(e_3)=0_W. $ Here $0_W$ is the zero output vector. In words: - The first input direction produces the first output direction. - The second input direction produces the output direction $f_2+f_3$. - The third input direction disappears: it produces zero. An arbitrary input is $ v=x_1e_1+x_2e_2+x_3e_3, $ where $x_1,x_2,x_3\in\mathbb R$ are its input coordinates. By linearity, $ Tv=x_1f_1+x_2(f_2+f_3)+x_3\,0_W =x_1f_1+x_2f_2+x_2f_3. $ Thus the coordinate calculation is $ \underbrace{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}_{[v]_{\mathcal B}:\ \text{input coordinates in the }e\text{-basis}} \xmapsto{\ A\ } \underbrace{\begin{pmatrix}x_1\\x_2\\x_2\end{pmatrix}}_{[Tv]_{\mathcal C}:\ \text{output coordinates in the }f\text{-basis}}. $ The arrow is labeled $A$, because it acts on coordinate columns. The underlying map $T$ acts on the actual vectors $v$ and $Tv$. - Question: The input is with respect to basis $\mathcal{B}$ and the output with respect to basis $\mathcal{C}$? - Yes. More precisely, the input column is $[v]_{\mathcal B}$ and the output column is $[Tv]_{\mathcal C}$. The actual input is $v=x_1e_1+x_2e_2+x_3e_3$; the actual output is $Tv=x_1f_1+x_2f_2+x_2f_3$. - What if I just wrote the output with respect to standard basis $e$? Wait a matrix is ALWAYS defined with respect to input output bases. - In this note, the $e_i$ belong to $V$, not $W$, so they are not the output basis. If domain and codomain were the same actual space, you could choose the same standard basis for both roles. A matrix representing a map still uses an input basis and an output basis; they may happen to be the same basis. Merely renaming the output basis is harmless, but silently switching which basis its entries refer to is not. ### Building its matrix from the map The columns of the matrix are the $\mathcal C$ output coordinates of the $\mathcal B$ input basis vectors: $ A= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&1&0 \end{pmatrix}. $ - Column 1 is $[T(e_1)]_{\mathcal C}=(1,0,0)^{\mathsf T}$. - Column 2 is $[T(e_2)]_{\mathcal C}=(0,1,1)^{\mathsf T}$. - Column 3 is $[T(e_3)]_{\mathcal C}=(0,0,0)^{\mathsf T}$. - The superscript $\mathsf T$ turns a displayed row of entries into a column; it is not the linear map $T$. Consequently, $ A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}x_1\\x_2\\x_2\end{pmatrix}. $ Here the column on the left is $[v]_{\mathcal B}$; the result on the right is $[Tv]_{\mathcal C}$. They use different bases because they describe vectors in different spaces. The columns show how input directions contribute to the output. The rows show formulas for individual output coordinates. These are two readings of the same multiplication. ### Checkpoint 1 — inputs versus outputs Let $v_1=2e_1+3e_2+7e_3$ and $v_2=2e_1+3e_2-4e_3$, both inputs in $V$. In your retained working below, $T(a,b,c)$ means $T(ae_1+be_2+ce_3)$: its argument is specified in the input basis $\mathcal B$. 1. Compute $Tv_1$, $Tv_2$, and their output coordinates in $\mathcal C$. What changed in the input? What changed in the output? - $T(2,3,7)=2f_{1}+3(f_{2}+f_{3})$ I.e. represented by coordinates $(2, 3, 3)$ with respect to basis $f$. - $T(2,3,-4)=(2f_{1},3(f_{2}+f_{3}), 0)$. i.e. represented by coordinate $(2,3,3)$ with respect to basis $f$. - Notation correction to the preceding line: the actual output is the sum $Tv_2=2f_1+3(f_2+f_3)$, not a tuple of vector-valued terms. Its coordinate column is $[Tv_2]_{\mathcal C}=(2,3,3)^{\mathsf T}$. The same is true for $Tv_1$. 2. Does having two different inputs with the same output violate the definition of a function? Which stronger property does it violate? - No, because two things can map to the same output. - The function is not injective. > [!tip]- Check your work > 1. Both actual outputs are $2f_1+3f_2+3f_3\in W$. Their output-coordinate columns are $[Tv_1]_{\mathcal C}=[Tv_2]_{\mathcal C}=(2,3,3)^{\mathsf T}$. Only the third input coordinate in $\mathcal B$ changed; the actual output did not change. > 2. No. Each input still has exactly one output. The map is not injective, because different inputs can have the same output. ## 4. Range, nullspace, and one target's solution set ### The range lives in the output space The range is $ \operatorname{range}T=\{Tv:v\in V\}\subseteq W. $ This means “collect the outputs of all inputs.” The symbol $\subseteq$ means “is a subset of.” For this map, $ \operatorname{range}T =\{p f_1+q f_2+q f_3:p,q\in\mathbb R\} =\operatorname{span}\{f_1,f_2+f_3\}. $ These are actual vectors in $W$. Their coordinate columns in the output basis $\mathcal C=(f_1,f_2,f_3)$ form the set $ \{[w]_{\mathcal C}:w\in\operatorname{range}T\} =\left\{\begin{pmatrix}p\\q\\q\end{pmatrix}:p,q\in\mathbb R\right\} =\operatorname{range}A. $ Thus the list $(p,q,q)$ describes an output in the $f$-basis, not an input in the $e$-basis. The vector itself is $p f_1+q f_2+q f_3$. - $p,q$: arbitrary real parameters describing reachable outputs. - $\operatorname{span}$: the set of all linear combinations of the listed vectors. - $f_1$ and $f_2+f_3$: two independent output directions. Therefore the range is a plane, of dimension $2$. In $\mathcal C$ output coordinates its equation is “third coordinate equals second coordinate,” not “third coordinate equals zero.” There are infinitely many outputs in this plane. Dimension $2$ means two independent directions, not two vectors. ### The nullspace lives in the input space The nullspace, also called the kernel, is $ \ker T=\{v\in V:Tv=0_W\}\subseteq V. $ This means “collect all inputs whose output is zero.” For the running example, $ T(x_1e_1+x_2e_2+x_3e_3)=0_W \quad\Longleftrightarrow\quad x_1=0,\quad x_2=0,\quad x_3\text{ arbitrary}. $ Hence $ \ker T=\{t e_3:t\in\mathbb R\} =\operatorname{span}\{e_3\}. $ These are actual input vectors in $V$. Their coordinate columns use the input basis $\mathcal B$, not the output basis $\mathcal C$: $ \{[v]_{\mathcal B}:v\in\ker T\} =\left\{\begin{pmatrix}0\\0\\t\end{pmatrix}:t\in\mathbb R\right\} =\ker A. $ Here $\ker A$ means the set of numerical input columns $x$ satisfying $Ax=0$. Its dimension is $1$: one independent input direction disappears. Every linear map sends the zero input to the zero output, so its nullspace is never empty. The distinction is: - $\ker T=\{0_V\}$: only the zero input maps to zero; $T$ is injective. - $\ker T\ne\{0_V\}$: a nonzero input also maps to zero; $T$ is not injective. Here $0_V$ denotes the zero input vector. There are not multiple different “zero vectors” in one space; there are multiple inputs producing its one zero output. ### A solution set is attached to one specified output Fix the reachable target $ w=2f_1+3f_2+3f_3\in W, \qquad [w]_{\mathcal C}=\begin{pmatrix}2\\3\\3\end{pmatrix}. $ Solving $Tv=w$ requires $x_1=2$ and $x_2=3$, but imposes no condition on $x_3$. Therefore its inputs are $ v=2e_1+3e_2+t e_3,\qquad t\in\mathbb R. $ In the input basis $\mathcal B$, the same solution vectors have coordinate columns $ [v]_{\mathcal B} =\begin{pmatrix}2\\3\\t\end{pmatrix} =\begin{pmatrix}2\\3\\0\end{pmatrix} +t\begin{pmatrix}0\\0\\1\end{pmatrix}. $ - $v_0=2e_1+3e_2$: one particular input producing $w$; its $\mathcal B$ coordinates are $(2,3,0)^{\mathsf T}$. - $t e_3$: an arbitrary nullspace vector; its $\mathcal B$ coordinates are $(0,0,t)^{\mathsf T}$. - ==The whole solution set: a line in the input space, not the range plane in the output space.== This is the general mechanism. Let $v_0$ be one particular input with $Tv_0=w$, and let $u$ be any nullspace vector. Then $ T(v_0+u)=Tv_0+Tu=w+0_W=w. $ Conversely, if $Tv=w=Tv_0$, then $T(v-v_0)=0_W$, so $v-v_0\in\ker T$. Thus all solutions, and only the solutions, form $ v_0+\ker T=\{v_0+u:u\in\ker T\}. $ ==This is a translated nullspace. It need not itself be a subspace: it generally does not contain the zero input.== The range answers “which targets can be hit?” The nullspace answers “which input changes leave the output unchanged?” Together they explain the solution set. ### Checkpoint 2 — two different spaces 1. Is the target $w=2f_1+3f_2+4f_3$, whose $\mathcal C$ coordinates are $(2,3,4)^{\mathsf T}$, in the range? How many inputs produce it? - None. Because $T(x_{1},x_{2},x_{3})=(x_{1}f_{1},x_{2}(f_{2}+f_{3}), 0)$. 2. For target $w=2f_1+3f_2+3f_3$, whose $\mathcal C$ coordinates are $(2,3,3)^{\mathsf T}$, identify the dimension of the range, the nullspace, and the solution set. Say which space contains each set. - W contians range $T$ is dim 2 - V contains nullspace of $T$ is dim 1, given by $(0,0,c)$ for all $c\in \mathbb{R}$ - Only inputs of the form (2,3, $t$) map to $w$. Hence it is a line in the codomain. It is a translate of the nullspace. Affine line in $V$. > [!tip]- Check your work > 1. No. Every output has equal second and third coordinates in $\mathcal C$, so there are no solutions. > 2. The range has dimension $2$ and lies in $W$, the output space. The nullspace has dimension $1$ and lies in $V$, the input space. This target's solution set is an affine line in $V$, also of dimension $1$: a translate of the nullspace. ## 5. Row reduction first: equivalent equations Before interpreting row operations as a basis change, read them as operations on equations. Let $b=[w]_{\mathcal C}=(b_1,b_2,b_3)^{\mathsf T}$ be the output-coordinate column of a target $w\in W$, and let $x=[v]_{\mathcal B}=(x_1,x_2,x_3)^{\mathsf T}$ be the **unknown** input-coordinate column. The equation $Ax=b$ means $ x_1=b_1,\qquad x_2=b_2,\qquad x_2=b_3. $ Subtracting the second equation from the third gives the equivalent system $ x_1=b_1,\qquad x_2=b_2,\qquad 0=b_3-b_2. $ No input solution was added or removed: adding the second equation back reverses the operation. In augmented-matrix notation, $ \left[\begin{array}{ccc|c} 1&0&0&b_1\\ 0&1&0&b_2\\ 0&1&0&b_3 \end{array}\right] \xrightarrow{\ \mathrm{row}_3\leftarrow\mathrm{row}_3-\mathrm{row}_2\ } \left[\begin{array}{ccc|c} 1&0&0&b_1\\ 0&1&0&b_2\\ 0&0&0&b_3-b_2 \end{array}\right]. $ - The vertical bar separates the coefficient matrix from the target column. - The coefficient matrix on the right is $R$, the reduced echelon form of $A$. - The last row says $0=b_3-b_2$. It tests whether the target is reachable. Thus $b_3=b_2$ gives infinitely many solutions, with $x_3$ free: it can be chosen arbitrarily. If $b_3\ne b_2$, there are no solutions. A system is consistent if it has at least one solution, and inconsistent if it has none. Consistent does not mean uniquely solvable. A nonzero bottom entry in the original target column is not the problem. A nonzero entry opposite a zero coefficient row after reduction is the problem. For example, $w=2f_1+3f_2+3f_3$ is reachable even though its third coordinate in $\mathcal C$ is $3$. ### Encoding that row operation as a matrix The operation is multiplication on the left by $ E=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&-1&1 \end{pmatrix}, \qquad E\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix} =\begin{pmatrix}y_1\\y_2\\y_3-y_2\end{pmatrix}. $ - $E$: the invertible matrix encoding the row operation. - $y_1,y_2,y_3$: entries of an arbitrary output-coordinate column $[w]_{\mathcal C}$, not new inputs to $T$. The result $E[w]_{\mathcal C}$ is its transformed column; Section 6 will identify it as $[w]_{\mathcal C'}$ in a new output basis. - Invertible means reversible: from $(y_1,y_2,y_3-y_2)$ we recover the old entries by adding the second entry to the third. We have $ R=EA=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&0 \end{pmatrix}, \qquad Ax=b\quad\Longleftrightarrow\quad Rx=Eb. $ The target column changes from $b=[w]_{\mathcal C}$ to $Eb$ along with the coefficient matrix. This does not require changing the actual target $w$: Section 6 interprets $Eb$ as its new output coordinates. In general, $Ax=b$ is not equivalent to $Rx=b$. ### Checkpoint 3 — transform the target too 1. Compute $Eb$ for targets with $\mathcal C$ output-coordinate columns $b=(2,3,3)^{\mathsf T}$ and $b=(2,3,4)^{\mathsf T}$. What does the last reduced equation say in each case? 2. For $w=f_2+f_3$, with $b=[w]_{\mathcal C}=(0,1,1)^{\mathsf T}$, compare $Ax=b$ with $Rx=b$. Why does comparing these two systems without changing $b$ give the wrong conclusion about row reduction? > [!tip]- Check your work > 1. The transformed output columns are $Eb=(2,3,0)^{\mathsf T}$ and $Eb=(2,3,1)^{\mathsf T}$. In Section 6 these are the targets' $\mathcal C'$ coordinates. Their last equations are $0=0$ and $0=1$, respectively. The first system has solutions; the second does not. > 2. $Ax=b$ has actual inputs $v=e_2+t e_3$ for every real $t$, with input-coordinate columns $x=[v]_{\mathcal B}=(0,1,t)^{\mathsf T}$. But $Rx=b$ has no solutions, because its last equation is $0=1$. The equivalent reduced system is $Rx=Eb$, where $Eb=(0,1,0)^{\mathsf T}$ is the transformed output column, not $Rx=b$. ## 6. Now interpret the same computation as a basis change Keep $T$, the spaces, and the input basis $\mathcal B$ fixed. Change only the output basis to $ \mathcal C'=(f_1,\ f_2+f_3,\ f_3). $ The prime on $\mathcal C'$ means “the new output basis.” These three vectors are independent and span $W$, so they are a basis of the whole codomain, not merely of the range. Take any vector in the codomain $W$ (a possible target, whether reachable or not): $ w=y_1f_1+y_2f_2+y_3f_3. $ Rewrite the same vector using the new basis: $ w=y_1f_1+y_2(f_2+f_3)+(y_3-y_2)f_3. $ Expanding the right-hand side recovers the original expression. Therefore $ [w]_{\mathcal C}=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}, \qquad [w]_{\mathcal C'}=\begin{pmatrix}y_1\\y_2\\y_3-y_2\end{pmatrix} =E[w]_{\mathcal C}. $ Nothing happened to $w$. Its coefficient list changed because the vectors used to describe it changed. ### One output, written twice For the input $e_2$, the actual output is $T(e_2)=f_2+f_3$. Its two coordinate columns are $ [T(e_2)]_{\mathcal C}=\begin{pmatrix}0\\1\\1\end{pmatrix}, \qquad [T(e_2)]_{\mathcal C'}=\begin{pmatrix}0\\1\\0\end{pmatrix}. $ The new column does not say the output became $f_2$. It says the output is one copy of the second new basis vector, which is $f_2+f_3$. In the new output basis $\mathcal C'$, the three images have coordinate columns $[T(e_1)]_{\mathcal C'}=(1,0,0)^{\mathsf T}$, $[T(e_2)]_{\mathcal C'}=(0,1,0)^{\mathsf T}$, and $[T(e_3)]_{\mathcal C'}=(0,0,0)^{\mathsf T}$. Those are precisely the columns of $R$. Thus $ \underbrace{R[v]_{\mathcal B}}_{\text{new output coordinates}} = \underbrace{[Tv]_{\mathcal C'}}_{\text{same output vector, new basis}}. $ ### The two interpretations must not be mixed | Interpretation | What does $R$ represent? | Did the actual output change? | | --- | --- | --- | | Keep the original output basis $\mathcal C$. | A different map, whose output for input $v=x_1e_1+x_2e_2+x_3e_3$ is $x_1f_1+x_2f_2$. Its output coordinates in $\mathcal C$ are $(x_1,x_2,0)^{\mathsf T}$. | Generally yes. | | Change the output basis from $\mathcal C$ to $\mathcal C'$. | The original map $T$, now described in new output coordinates. | No. | Both interpretations are valid. Saying only “the same map” without specifying the bases leaves out essential information. ### What happens to range and nullspace? The actual range of $T$ remains $ \operatorname{span}\{f_1,f_2+f_3\}\subseteq W. $ The two coordinate descriptions of that same range are $ \{[w]_{\mathcal C}:w\in\operatorname{range}T\} =\left\{\begin{pmatrix}p\\q\\q\end{pmatrix}:p,q\in\mathbb R\right\} =\operatorname{range}A, $ $ \{[w]_{\mathcal C'}:w\in\operatorname{range}T\} =\left\{\begin{pmatrix}p\\q\\0\end{pmatrix}:p,q\in\mathbb R\right\} =\operatorname{range}R. $ The second description does not mean the actual plane moved. The new third coordinate measures the coefficient of $f_3$ after using $f_2+f_3$ as the second basis vector. The input basis never changed. In fact, as sets of input-coordinate columns, $ \ker R=\ker A=\{[v]_{\mathcal B}:v\in\ker T\}, $ because $Rx=0$ means $EAx=0$, and invertibility of $E$ makes that equivalent to $Ax=0$. For the ranges of the numerical matrices, the precise relationship is $ \operatorname{range}R=E(\operatorname{range}A). $ These need not be the same subsets of coordinate space, but have the same dimension. Under the basis-change interpretation, they are two coordinate descriptions of the same underlying range of $T$. ### Checkpoint 4 — what does a coordinate mean? 1. A vector has new output coordinates $[w]_{\mathcal C'}=(0,1,0)^{\mathsf T}$. Which actual vector in $W$ do they describe? Contrast this with interpreting the same list as old output coordinates $[w]_{\mathcal C}$ instead. 2. For the actual target $w=f_2$, find its old and new coordinates. Is it in the range of $T$? > [!tip]- Check your work > 1. $[w]_{\mathcal C'}=(0,1,0)^{\mathsf T}$ describes $w=f_2+f_3$. The same numerical list interpreted in $\mathcal C$ describes the different vector $f_2$. The list alone does not specify the vector unless the basis is understood. > 2. $[f_2]_{\mathcal C}=(0,1,0)^{\mathsf T}$ and $[f_2]_{\mathcal C'}=(0,1,-1)^{\mathsf T}$. It is outside the range: its $\mathcal C$ coordinates do not have equal second and third entries, and its $\mathcal C'$ third coordinate is nonzero. These are the same obstruction in different coordinates. ## 7. Translate this into pivots, rank, and free variables Now let $A$ be any $m\times n$ real matrix. - $n$: number of columns, hence the dimension of the input-coordinate space $\mathbb R^n$. - $m$: number of rows, hence the dimension of the output-coordinate space $\mathbb R^m$. - $r$: number of pivots in echelon form, called the rank of $A$. - A pivot entry: the leading nonzero entry of a nonzero echelon row. In reduced echelon form it equals $1$ and is the only nonzero entry in its column. The bridge is: | Matrix language | Map language | | --- | --- | | $r$ pivots. | The range has dimension $r$. | | $n-r$ free variables. | The nullspace has dimension $n-r$. | | $r=n$: a pivot in every column. | The map is injective: its nullspace is $\{0\}$. | | $r=m$: a pivot in every row. | The map is surjective (onto): every target in the codomain is reachable. | | A contradiction row $[0\ \cdots\ 0\mid c]$, $c\ne0$. | The target is outside the range. | | A consistent system with no free variables. | Exactly one input reaches this target. | | A consistent system with at least one free variable. | A positive-dimensional translated nullspace reaches this target; over $\mathbb R$, there are infinitely many solutions. | For the running example, $n=m=3$ and $r=2$. Rank–nullity says $ \underbrace{3}_{\dim V} = \underbrace{2}_{\dim\operatorname{range}T} + \underbrace{1}_{\dim\ker T}. $ Here two independent input directions have independent output images, and the remaining direction lies in the kernel. Such a choice of input directions is possible in general; it need not coincide with the original coordinate axes. Three cautions: 1. Count pivots, not merely nonzero rows of an unreduced matrix. Our original $A$ had three nonzero rows but only two independent rows and two pivots. 2. A zero coefficient row expresses a restriction on reachable targets. A free variable concerns undetermined input coordinates. For square $3\times3$ matrices they occur together when rank drops, but they are not the same concept. 3. To obtain a basis for the original matrix's column space from pivot positions, use the corresponding columns of the original matrix, not the row-reduced columns. Row operations generally change the column space as a subset of fixed output coordinates. ### Checkpoint 5 — does a zero row force infinitely many solutions? Consider the map $S:\mathbb R\to\mathbb R^2$ with matrix $ N=\begin{pmatrix}1\\0\end{pmatrix}, \qquad S(t)=t g_1, \qquad [S(t)]_{\mathcal D}=\begin{pmatrix}t\\0\end{pmatrix}. $ Here the input basis of $\mathbb R$ is $(1)$, and $\mathcal D=(g_1,g_2)$ is the standard output basis of $\mathbb R^2$. The basis vector $g_1$ is the horizontal direction and $g_2$ the vertical direction. We use new letters because this example has a two-dimensional codomain, unlike the running example. The scalar $t$ is a single real input, and $N$ has one column and two rows. 1. Find the range, nullspace, and number of solutions for target $w=2g_1$, with output coordinates $[w]_{\mathcal D}=(2,0)^{\mathsf T}$. Does the zero row make $S$ noninjective? 2. How many inputs reach $w=2g_1+g_2$, with output coordinates $[w]_{\mathcal D}=(2,1)^{\mathsf T}$? What does the zero row tell you instead? > [!tip]- Check your work > 1. The actual range is $\{t g_1:t\in\mathbb R\}=\operatorname{span}\{g_1\}$, of dimension $1$, in the output space $\mathbb R^2$. The nullspace is $\{0\}$, of dimension $0$, in the input space $\mathbb R$. The unique input reaching $w=2g_1$ is $t=2$. The map is injective despite its zero row: there is one pivot for its one input variable. > 2. There are no solutions. The zero row says every reachable target must have second output coordinate zero in $\mathcal D$. There are fewer reachable outputs, but no input freedom for the outputs that are reachable. ## 8. Transfer back to the earlier matrix Return to the example in [[05 Library/Linear Algebra Done Wrong/2 Reduction and Echelon/2B5 - Reduced Echelon as Linear Maps.md|2B5 - Reduced Echelon as Linear Maps]]. This time the third input direction does not disappear on its own: a combination of input directions disappears. Use the original spaces $V,W$ and their original bases $\mathcal B=(e_1,e_2,e_3)$ and $\mathcal C=(f_1,f_2,f_3)$. Let $ M=\begin{pmatrix} 1&2&3\\ 0&1&2\\ 0&0&0 \end{pmatrix}. $ The actual map is $ U(x_1e_1+x_2e_2+x_3e_3) =(x_1+2x_2+3x_3)f_1+(x_2+2x_3)f_2. $ Equivalently, its coordinate calculation is $ M\underbrace{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}_{[v]_{\mathcal B}} =\underbrace{\begin{pmatrix}x_1+2x_2+3x_3\\x_2+2x_3\\0\end{pmatrix}}_{[Uv]_{\mathcal C}}. $ - $U:V\to W$: the actual map between the two copies of $\mathbb R^3$. - $M$: its matrix using input basis $\mathcal B$ and output basis $\mathcal C$. - $x_1,x_2,x_3$: input coordinates in $\mathcal B$, not output coordinates. ### Checkpoint 6 — put the bridge to work 1. Find a nonzero vector in $\ker U$. Find the range and the dimensions of the range and nullspace. Why does having no zero column not imply injectivity? 2. Solve $Uv=w$ for target $w=f_1-f_2$, whose output coordinates are $[w]_{\mathcal C}=(1,-1,0)^{\mathsf T}$. Write the answer as one particular input plus an arbitrary nullspace vector, and then give its input coordinates in $\mathcal B$. 3. Replace the target with $w=f_1-f_2-4f_3$, whose output coordinates are $[w]_{\mathcal C}=(1,-1,-4)^{\mathsf T}$. What changes? What is the last augmented equation? > [!tip]- Check your work > 1. $U(e_1-2e_2+e_3)=0_W$. Reducing gives $x_1-x_3=0$ and $x_2+2x_3=0$, so the actual nullspace is $\ker U=\operatorname{span}\{e_1-2e_2+e_3\}\subseteq V$, of dimension $1$. The generator's input-coordinate column is $(1,-2,1)^{\mathsf T}$ in $\mathcal B$. The actual range is $\{p f_1+q f_2:p,q\in\mathbb R\}=\operatorname{span}\{f_1,f_2\}\subseteq W$, of dimension $2$; its $\mathcal C$ coordinate columns have the form $(p,q,0)^{\mathsf T}$. The first two columns of $M$ span that coordinate plane. No single input basis vector disappears, but a nontrivial combination does. The columns are dependent. > 2. Subtract twice row 2 from row 1, applying this to the target column $[w]_{\mathcal C}$ too. The equations become $x_1-x_3=3$, $x_2+2x_3=-1$, and $0=0$. Thus the actual inputs are $v=(3e_1-e_2)+t(e_1-2e_2+e_3)$ for every real $t$. Their input-coordinate columns are $[v]_{\mathcal B}=(3,-1,0)^{\mathsf T}+t(1,-2,1)^{\mathsf T}$. The first vector is a particular input; the second term ranges over the nullspace. > 3. The last equation becomes $0=-4$. There are no solutions: the target is outside the range. The nontrivial nullspace does not make an unreachable target reachable. ## 9. The compact picture to keep - The map sends inputs in $V$ to outputs in $W$. A matrix performs that operation on coordinate columns after bases are chosen. - Range: which outputs occur. Nullspace: which input changes produce no output change. - A reachable target has solution set “one particular input $+$ nullspace.” An unreachable target has no solutions. - Row reduction preserves solutions only when the target column is transformed too: $Ax=b\iff Rx=Eb$. - $R$ describes the same abstract map only when the changed output coordinates are understood. In fixed standard coordinates, it generally describes a different map. - Rank and nullity count independent directions. For a matrix with $n$ columns, their sum is $n$, the input dimension.