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I am so lost on this question I am not sure even where to start. I am not looking for an answer but more of a turn in the right direction.

Find all the eigenvalues and eigenfunctions of the following linear transformation: $T(f(x)) = f(-x)$ from $\mathbf{P}_2$ to $\mathbf{P}_2$. IS T diagonalizable?

All I know for sure is that the standard basis is $1$, $x$, $x^2$, and if this is an eigenbasis for $T$ then $T$ is diagonalizable.

Thanks for any assistance!

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What is $P_2$? The space of polynomials with real coefficients and degree at most 2? Something else? –  Andres Caicedo Jan 5 '11 at 1:14

3 Answers 3

An eigenvector of a linear tranformation $T\colon\mathbf{V}\to\mathbf{V}$ is a vector $\mathbf{v}$, $\mathbf{v}\neq\mathbf{0}$, such that there is a scalar $\lambda$ for which $T(\mathbf{v})=\lambda \mathbf{v}$. The scalar $\lambda$ is called an eigenvalue. $T$ is diagonalizable if and only if you can find a basis for $\mathbf{V}$ in which every vector in the basis is an eigenvector. This does not mean that any basis will consist of eigenvectors. It means that there is a way to pick a basis so that every element of the basis is an eigenvector.

So, here, $\mathbf{V}=\mathbf{P}_2$. Your function $T$ is defined by the given formula. For example, if $p(x) = x^2 - 3x + 1$, then $p(-x) = (-x)^2 -3(-x) + 1 = x^2 +3x + 1$, so $T(p) = x^2+3x+1$.

You are looking for eigenvalues and eigenvectors. If you already know about the characteristic polynomial, then you should figure out the characteristic polynomial of $T$ (say, using the standard basis to find a coordinate matrix for $T$) and proceed along those lines.

If you don't know about the characteristic polynomial, not all is lost. Suppose $p(x) = ax^2+bx+c$ is an eigenvector. That means that $a$, $b$, and $c$ are not all zero, and there is a scalar $\lambda$ such that $T(p)=\lambda p$. Since $T(p(x)) = ax^2 -bx +c$ and $\lambda p = \lambda ax^2 + \lambda bx + \lambda c$, the equation says that $$ax^2 - bx + c = T(p) = \lambda p = \lambda ax^2 +\lambda bx + \lambda c.$$ For this to happen, you must have: \begin{align*} a &= \lambda a\\ -b &= \lambda b\\ c &=\lambda c \end{align*} and $a$, $b$, and $c$ not all zero. You want to find all values of $\lambda$ for which there are soluctions $a,b,c$ not all zero. This is not terribly hard to do. Try that, and we can take it from there; edit your question if you get stuck to account for how far you have managed to go then.

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Arturo: Why not argue from the standard basis, to find the eigenvalues (and eigenfunctions) directly? –  Andres Caicedo Jan 5 '11 at 1:25
    
@Andres: Sure, in this particular situation that would work; I was talking more about how to go about it in general. –  Arturo Magidin Jan 5 '11 at 1:42
    
I had a comment earlier about how diagonalizability only requires the existence of some basis of eigenfunctions, not that the standard basis would work, but I erased since the phrasing of the question made me think it would be unnecessarily confusing to the OP. –  Andres Caicedo Jan 5 '11 at 2:09

Here's a general remark, which may or may not be helpful to you:

The transformation $T$ in your question has the property that if you do it twice you get back to what you started with. Expressed differently, $T$ composed with itself is the identity operator: $T \circ T = I$. (This is because $-(-x)=x$; changing the sign of the variable $x$ twice is the same as doing nothing to $x$.) Such a linear transformation is called an involution, and involutions have the property that the eigenvalues can only be $+1$ and $-1$. This is proved as follows: suppose $v$ is an eigenvector with eigenvalue $\lambda$; this implies that $$T(T(v))=T(\lambda v)=\lambda T(v) = \lambda \lambda v = \lambda^2 v,$$ and moreover $T(T(v))=v$ since $T$ is an involution. These two expressions for $T(T(v))$ must be equal, so $\lambda^2 v=v$, and since an eigenvector by definition is nonzero, it follows that $\lambda^2 = 1$. Hence $\lambda=\pm 1$. Q.E.D.

If $T$ is an involution, the following also holds: for any vector $u$, the vectors $u \pm T(u)$ are eigenvectors (if they are nonzero) with eigenvalues $\pm 1$. (Proving this is a nice little one-line exercise.)

This question is somewhat related.

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The conventional way to solve the problem goes as follows:

Let's begin with the standard basis of $\mathbf{P}_2$ viz {$1,x,x^2$}. $$\begin{matrix}T(1)&=&1&=&1\times 1&+&0 \times x&+&0\times x^2\\ T(x)&=&-x&=&0\times 1&+&(-1) \times x&+&0\times x^2\\ T(x^2)&=&x^2&=&0\times 1&+&0 \times x&+&1\times x^2\end{matrix}$$ $\Rightarrow$ Mat $T$ = $ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ $\Rightarrow$ T is evidently diagonalizable & Eigen values of T are $1, -1, -1$. Calculating eigen vectors is just a matter of calculation e.g. the e.vector corresponding to the e.value $-1$ can be calculated as: $ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$ \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix}$ =$-1$$ \begin{pmatrix} x_1\\ x_2\\ x_3\\ \end{pmatrix} \Rightarrow$ $\begin{pmatrix} x_1\\ -x_2\\ x_3\\ \end{pmatrix} = \begin{pmatrix} -x_1\\ -x_2\\ -x_3\\ \end{pmatrix}\Rightarrow${$(0,c,0)^t:c\in\mathbb R$} is the eigenspace of $-1$. Similar arguments yield the other eigen spaces.

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