Verify, whether matrices form a basis for $\mathbb{R}_{2\times 2}$ and find coordinates in the basis

Question

I'm trying the solve the following problem:

Verify, whether matrices $A=\begin{pmatrix} -1&1 \\ 1&2 \end{pmatrix}$, $B=\begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}$, $C= \begin{pmatrix} 1&1 \\ 0&0 \end{pmatrix}$ and $D=\begin{pmatrix} 0&0 \\ 1&2 \end{pmatrix}$ form a basis for $\mathbb{R}_{2\times 2}$. Also find coordinates of matrix $E=\begin{pmatrix} -2&2 \\ 8&4 \end{pmatrix}$ in the basis.

What I did was to create a matrix $F=$ $\begin{pmatrix} -1&0 &1 &0 \\ 1&1 &0 &1 \\ 1&1 &1 &0 \\ 2&0 &0 &2 \end{pmatrix}$ and verify, whether there are any linearly dependent lines. I found out that there are none, therefore the matrices $A,B,C,D$ are generating the entire space and therefore they form a basis. Is my solution correct?

And how would I find the coordinates of the matrix E in the basis? Do I just put the numbers from the matrix $E$ as solutions in my matrix $F$? Thanks!

Answer 1

You found they form a basis. Now you need to solve $aA+bB+cC+dD=E$ for $a,b,c$ and $d$. This is a linear system of $4$ equations in $4$ unknowns. You could form another matrix to solve.

Here's the matrix: $\left(\begin {array}{rrrr|r}-1&0&1&0&-2\\1&1&1&0&2\\1&1&0&1&8\\2&0&0&2&4\end{array}\right)$.

Answer 2

If you have a basis $B = \{v_1, v_2, v_3, \dots\}$ and a corresponding matrix $M = [v_1, v_2, v_3, \dots]$, both expressed in the "standard" basis, then the inverse of this matrix is the change-of-basis matrix from the standard basis to $B$. So, you just need to find the inverse of the matrix $F$ and multiply by $\begin{bmatrix} -2 \\ 8 \\ 2 \\ 4 \end{bmatrix}$. This will give you a column vector version of the matrix in the correct basis.

Answer 3

Your method for proving that $A,B,C,D$ form a basis for the set of matrices $\Bbb{R}_{2 \times 2}$ is correct, but perhaps you could benefit from a little more justification. What you have done is to "stack the matrix as a column vector". More precisely, this means you have constructed a linear isomorphsim $\Phi: \Bbb{R}_{2 \times 2} \to \Bbb{R}^4$ defined by the rule \begin{align} \Phi \begin{pmatrix} a & c \\ b & d \end{pmatrix} &= \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \end{align}

You were asked to verify whether the set \begin{equation} \alpha = \{A,B,C,D\} \end{equation} forms a basis for $\Bbb{R}_{2 \times 2}$. Since $\Phi$ is an isomorphism, this is equivalent to checking whether \begin{equation} \Phi[\alpha] = \{\Phi(A), \Phi(B), \Phi(C), \Phi(D)\} \end{equation}

forms a basis of $\Bbb{R}^4$. What you have done next is to create a new matrix $F$ whose columns are $\Phi(A), \dots \Phi(D)$. After this, you verified that these four columns are linearly independent. Thus, you found $4$ linearly independent vectors in $\Bbb{R}^4$, which means they form a basis. Therefore, $\alpha = \{A,B,C,D\}$ does indeed form a basis of $\Bbb{R}_{2 \times 2}$.

This is a more complete explanantion of why your method of "stacking matrices as column vectors" and checking whether the $4$ columns are independent/dependent actually works.