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I have this question in my textbook:

  1. Find a basis B of $R^2$ such that the matrix of the linear transformation $T(x, y) = (y, x)$ is diagonal with respect to B, and give the diagonal matrix

I have no idea how to proceed.

This is the only relevant explanation in my textbook:

some image description

Maybe I can ask, how did they get that diagonal matrix in the book from the basis?

I can see how they got the standard matrix.

and I can see how they got the basis:

enter image description here

But in the first photo, how do they go from the basis to the diagonal matrix?

EDIT

Is this right?

I have A =

$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

makes sense to me that this is the linear transformation matrix since

$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} * \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$

So finding eigenvalues:

$$(\lambda + 1)(\lambda - 1)$$

so eigenvalues are (1, -1)

Finding eigenvectors:

$$I - A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$

$$-> \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} t \\ t \end{bmatrix} = t * \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Doing the same thing for eigenvalue = -1 leads me to eigenvector = $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$

So $P = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$

And $$P^{-1} = \begin{bmatrix} 1 & -1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{bmatrix}$$

$$ = \begin{bmatrix} 1 & -1 & 1 & 0 \\ 0 & 2 & -1 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & \frac{-1}{2} & \frac{1}{2} \end{bmatrix}$$

so $$P^{-1} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{1}{2} \end{bmatrix}$$

I get the diagonal matrix defined by $$P^{-1} * A * P = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$

Kitty Capital
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2 Answers2

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First you should find the matrix that represents the linear transformation $T$ with respect to the standard basis $\alpha$ which is: $ [T]_\alpha = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, since $\begin{bmatrix} 1 \\ 0 \end{bmatrix} $ maps to $\begin{bmatrix} 0 \\ 1 \end{bmatrix} $ and vice versa.

Next, you should try to diagonalize (get into the form $ [T]_\alpha =PDP^{-1}$) the matrix by finding its eigenvalues and eigenvectors. The eigenvectors will form a basis iff $[T]_\alpha$ is diagonalizable, since they are linearly independent and the quantity of them matches the vector space's dimension.

Additionally, they will be the column vectors of $P$ with column $P_i$ being the eigenvector for eigenvalue $\lambda_i$ in the column $D_i$. Call this basis of eigenvectors $\beta$.

Note that is because $P$ is the same as the change-of-basis matrix from $\beta$ to $\alpha$. Therefore, inverting $P$ will give you the change-of-basis matrix from $\alpha$ to $\beta$.

Intuitively, you can think of what happens when you pass a vector $\vec{x}$ in its standard representation into the matrices $PDP^{-1}$ as this: $P^{-1}$ represents $\vec{x}$ in terms of the new basis $\beta$ which is the basis we need in order to easily evaluate the function $T$ (multiplying a vector by a diagonal matrix). After those two multiplications, all we need is to multiply by $P$ to change our value back to the basis that we usually like to work with.

Anthony Ter
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  • Using A = $\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$, I get that the eignvalue is 1 of multiplicty 2, and the eigenvector is $\begin{bmatrix} 1 \ 1 \end{bmatrix}$. Where do I go from here? – Kitty Capital Sep 25 '18 at 19:00
  • If that were the case, nowhere. You only have one eigenvector so you can't make a basis for $\mathbb{R}^2$. Luckily, you made a simple mistake calculating eigenvalues and the actual characteristic polynomial is $t^2 - 1 = 0$, so eigenvalues are $-1, 1$. Once you have the new eigenvectors, make the $P$ matrix using the eigenvectors as columns. – Anthony Ter Sep 25 '18 at 19:09
  • I got the new eigenvectors to be (1,1), and (-1,1). Is this the basis? – Kitty Capital Sep 25 '18 at 19:16
  • I added some work to my original post @Anthony Ter. Mind taking a look? – Kitty Capital Sep 25 '18 at 19:46
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If you have a basis of eigenvectors $B$ for a linear operator $T$, then $[T]_B$ is diagonal. The converse is true too: if $[T]_B$ is diagonal for a basis $B$, then $B$ consists only of eigenvectors.

The proof follows directly from the definition of $[T]_B$. Suppose $B = (v_1, \ldots, v_n)$ are eigenvectors, with respective eigenvalues $\lambda_1, \ldots, \lambda_n$. Then, we compute $[T]_B$ by transforming each $v_i$ by $T$, and expressing the result in terms of coordinate column vectors with respect to $B$. These column vectors form the columns of $[T]_B$.

So, if we apply $T$ to $v_i$, we get $$T v_i = \lambda_i v_i = 0 v_1 + \ldots + 0 v_{i-1} + \lambda_i v_i + 0 v_{i+1} + \ldots + 0v_n,$$ hence the $i$th column of $[T]_B$ contains only $0$s except (possibly) in the $i$th position (i.e. on the main diagonal), where it is equal to $\lambda_i$. That is, $$[T]_B = \operatorname{diag}(\lambda_1, \ldots, \lambda_n).$$

So, if you follow how to find the basis $B$ of eigenvectors, then simply computing the matrix for $T$ with respect to this basis will quickly and easily lead you to the answer.

Theo Bendit
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