**7.1. True or false:** **(a)** The rank of a matrix is equal to the number of its non-zero columns; **(b)** The $m \times n$ zero matrix is the only $m \times n$ matrix having rank $0$; **(c)** Elementary row operations preserve rank; **(d)** Elementary column operations do not necessarily preserve rank; **(e)** The rank of a matrix is equal to the maximum number of linearly independent columns in the matrix; **(f)** The rank of a matrix is equal to the maximum number of linearly independent rows in the matrix; **(g)** The rank of an $n \times n$ matrix is at most $n$; **(h)** An $n \times n$ matrix having rank $n$ is invertible. **(a)** False **(b)** True **(c)** True **(d)** False. $A^{\top}$ and perform, its the same. **(e)** True. It is the count of LI columns or LI rows, but number of LI columns = number of LI rows. **(f)** True. It is the count of LI columns or LI rows, same as $e$. - Remember rank of $A$ comes about by the fact column rank and row rank equate. With $n$ column vectors of length $m$. If $n$ exceeds $m$, they can't be LI. So $\text{col rank A}\leq{}m$. Additionally, With $m$ row vectors of length $n$. If $m$ exceeds $n$, they can't be LI. So $\text{row rank }A\leq{}n$. By default we already know $\text{row rank }A\leq{}m$ and $\text{col rank}\leq{}n$. So $\text{row rank }A\text{, col rank }A\leq{}\min\{ m,n \}$. Whats the bridge for equating? - Rank of each col and row is already at most $\min{m,n}$. If a column doesn't have a pivot, then a row loses a pivot position. **(g)** True. **(h)** It'd be injective/bijective operator, so yes. **7.2.** A $54 \times 37$ matrix has rank $31$. What are dimensions of all $4$ fundamental subspaces? The matrix of of $T:\mathbb{F}^{37}\to \mathbb{F}^{54}$. 37 = dim kernel of $T$ + dim range T. - $\text{dim range }A=31$ - $\text{dim ker }A=6$ - $\text{dim range }A^{\top}=31$ - $\text{dim ker }A^{\top}=23$ Since $Tv_{1},\ldots,Tv_{31}\in \mathbb{F}^{54}$ are LI in $\mathbb{F}^{54}$ using only 37 coefficients per vector. $A^{T}_{37\times 54}$ is the map $S:\mathbb{F}^{54}\to \mathbb{F}^{37}$. So there are $31$ LI rows, so rank of $A^{T}=31$. So kernel $A^{T}=54-31=23$