# Diagonalization and linear transformations

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

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.

• 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

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.