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

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!

• Aren't you supposed to find if the given matrices form the basis for $\mathbb{R}_{2}$? Jul 1, 2019 at 16:42
• @InMyOwnEuclideanspace no, it's $\mathbb{R}_{2\times 2}$ Jul 1, 2019 at 16:44
• I think you might have mistakenly assumed that you need to find the basis for a four dimensional space, $\mathbb{R}_{2x2}$ does not mean that. What you need to do is figure out if each of the A,B,C and D matrices mentioned form a basis for $\mathbb{R}^{2}$. Jul 1, 2019 at 16:50
• @InMyOwnEuclideanspace How would I do so? Sorry, it says $\mathbb{R}_{2\times 2}$ in the problem's description and I have never seen a problem like this, so I'm quite lost in this. Jul 1, 2019 at 16:52
• @InMyOwnEuclideanspace The matrices are not even elements of $\Bbb{R}^2$, so how an they form a basis? I think the OP's answer is correct (it is missing some details regarding the isomorphism he implicitly used between $\Bbb{R}_{2 \times 2}$ and $\Bbb{R}^4$) but the idea is correct. Jul 1, 2019 at 16:53

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

• Thanks! Could you please show me an example how will the matrix look like? Do I just put the numbers from $E$ as a solution of my matrix $F$? Jul 1, 2019 at 16:58
• @james See my edit.
– user403337
Jul 1, 2019 at 17:05
• Thanks! One last question, why did you put $(-2,2,8,4)$ as a solution instead of $(-2,8,2,4)$? Does it matter whether I go put them by the columns or by the rows? Jul 1, 2019 at 17:05
• I did it in a different order than you. But the order doesn't matter. As long as you're consistent. (You could switch the $2$nd and $3$rd row.)
– user403337
Jul 1, 2019 at 17:11

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.

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.