Krdytkyu

Given matrix $$A = begin{pmatrix} 1 & r & 1 & 1 \ 1 & s & 1 & 2 \ 1 & t & 1 & 3 \ 1&0&0&1end{pmatrix}$$ and let $B$ be the matrix obtained from $A$ by exchanging the $2^{nd}$ and $4^{th}$ columns, i.e. $$B = begin{pmatrix} 1&1&1&r\1&2&1&s\1&3&1&t\1&1&0&0 end{pmatrix}$$

(i) Justify whether the column space of $A$ the same as the column space of $B$.

(ii)Justify whether the row spaces of $A$ and $B$ have the same dimension.

(iii) Let $C$ by an invertible 4-by-4 matrix. Determine a basis for the row space of the 4-by-8 matrix $$D = begin{bmatrix} mathbf{C} & mathbf{C}end{bmatrix}$$

For part (i), this is my attempt:

I note that $B^T$ is the matrix obtained from $A^T$ by swapping the $2^{nd}$ and $4^{th}$ rows of $A^T$, thus $B^T$ is row equivalent to $A^T$ since elementary row operations preserve the row space $-$ $A$ has the same column space as $B$.

For part (ii), I base my attempt on my answer from (i). $A$ and $B$ have the same column space, so $A$ and $B$ must have the same dimension of column space. Since $$dimRan(A) = dimRan(A^T)$$ and $$dimRan(B) = dimRan(A)$$ we get that $dimRan(B^T) = dimRan(A^T)$ i.e. the row spaces of $A$ and $B$ indeed have the same dimension.

For part (iii), I make use of the fact that $$rref(C) = begin{pmatrix} 1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1end{pmatrix}$$ so $$rref(D) = begin{bmatrix} 1&0&0&0&1&0&0&0\0&1&0&0&0&1&0&0\0&0&1&0&0&0&1&0\0&0&0&1&0&0&0&1end{bmatrix}$$

Thus we can take a basis for the row space of $D$ to be $${begin{pmatrix}1\0\0\0\1\0\0\0end{pmatrix},begin{pmatrix}0\1\0\0\0\1\0\0end{pmatrix},begin{pmatrix}0\0\1\0\0\0\1\0end{pmatrix},begin{pmatrix}0\0\0\1\0\0\0\1end{pmatrix}}$$

I'm pretty sure my answer for (i) is correct, but I'd just like to verify and see if there would be another way to explain it, but I'm unsure about part (ii) as intuitively, I would expect the row spaces of $A$ and $B$ to have different dimensions since $A$ is not row-equivalent to $B$.

My approach for part (iii) seems a little too naive and I feel like there's more to the question than meets the eye.

Help would be appreciated!

Seems that you know what to do.

For part $(i)$, the column space are spanned by the columns. Hence interchanging the columns doesn't change the space. Hence the column space is the same.

For part $(ii)$, the dimension of the row space is equal to the dimension of the column space. Hence the dimension of the row space is equal.

For part $(iii)$, if $P$ is invertible $PA$ and $A$ share the same row space. Hence for $D=[C, C]$. We can premultiply $D$ by $C^{-1}$ and obtain $[I, I]$. Hence a basis would be ${(e_i + e_{4+i})^T|i in { 1,2,3,4}}$ where $e_i in mathbb{R}^{8 times 1}$ is the $i$-th standard unit basis.