# Find the Change of Coordinate Matrix from Basis of $\mathbb{R}^2$ to Standard Basis

I'm not sure if I'm solving the following problem correctly so I would really appreciate if someone could help me out here.

Given a matrix $A = \begin{bmatrix}-37 & 24\\-60 & 39\end{bmatrix}$,

and $B = \{[2,3] , [3,5]\}$ is a basis of $\mathbb{R}^2$ consisting of eigenvectors for $A$.

I need to find the change of coordinate matrix $P = _S P_B$ where $S$ is the standard basis.

This is what I'm thinking I'm supposed to do but I'm really unsure. Can anyone let me know if I'm on the right track?

$\begin{bmatrix}1 & 0\end{bmatrix}$ $x_1$ + $\begin{bmatrix}0 & 1\end{bmatrix}$ $x_2$ = $\begin{bmatrix}-37 & 24\\-60 & 39\end{bmatrix}$

• No computation here: it is simply the matrix with columns equal to the coordinates of the eigenvectors in the canonical basis Dec 1, 2015 at 0:20
• I calculated the eigenvectors $\begin{bmatrix}0 & -1 & 0\end{bmatrix}$ and $\begin{bmatrix}1 & 0 & 0\end{bmatrix}$ and I'm sorry I'm a little confused about what you mean? Dec 1, 2015 at 0:30
• Would the inverse for a matrix containing the two columns of B give me the change of coordinates? Dec 1, 2015 at 0:31
• It lets you express the old (= in the canonical basis) coordinates in function of the new coordinates. Dec 1, 2015 at 0:39
• Oh! So I'm doing this wrong? Dec 1, 2015 at 0:45

There's, I'll adjust what @Bernard said slightly, almost no computation here. Well, it depends on whether you know an all-important secret or two.

There's the transition matrix, and on the other hand the change of basis matrix. They're inverses. (I tend to forget which is which).

But the matrix in this case is $$\begin{pmatrix}2\quad3\\3\quad 5\end{pmatrix}$$.

The reason is that it takes vectors expressed in terms of the basis consisting of the columns, and expresses them in the standard basis. This can be easily checked, by applying the matrix to the elements of the basis consisting in the columns, expressed in terms of itself. That means applying it to $$\begin {pmatrix}1\\0\end {pmatrix}$$ and $$\begin{pmatrix}0\\1\end{pmatrix}$$. And as you see out pop the columns, in the standard basis.

Now, back to what @Bernard said. Often one will need to go in the other direction. Thus you do need to compute the inverse of this matrix.

The answer is based on a best guess at the original question.

Given the standard basis $$\mathcal{S} = \left\{ \left[ \begin{array}{c} 1 \\ 0 \end{array} \right], % \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \right\}$$ the matrix $\mathbf{A}$ has the following representation: $$\left[ \mathbf{A} \right]_{\mathcal{S}} = \left[ \begin{array}{rc} -37 & 24 \\ -60 & 39 \\ \end{array} \right].$$

Given the basis of eigenvectors for $\mathbf{A}$, $$\mathcal{S}' = \left\{ \left[ \begin{array}{c} 3 \\ 5 \end{array} \right], % \left[ \begin{array}{c} 2 \\ 3 \end{array} \right] \right\},$$ express the matrix in this new basis.

Define the matrix $$\mathbf{Q} = \left[ \begin{array}{cc} 3 & 2 \\ 5 & 3 \\ \end{array} \right],$$ the solution is $$\left[ \mathbf{A} \right]_{\mathcal{S}'} = \mathbf{Q}^{-1} \left[ \mathbf{A} \right]_{\mathcal{S}} \mathbf{Q} = \left[ \begin{array}{r} -3 & 2 \\ 5 & -3 \end{array} \right] % \left[ \begin{array}{rr} -37 & 24 \\ -60 & 39 \end{array} \right] % \left[ \begin{array}{cc} 3 & 2 \\ 5 & 3 \end{array} \right] = \left[ \begin{array}{cr} 3 & 0 \\ 0 & -1 \end{array} \right].$$

This is the expected result: the eigenvectors are used to diagonalize a matrix.