# Write $V=P_2(\mathbb{R})$ as a direct sum of $V=W_1\oplus W_2 \oplus W_3$

So, if I let $T:P_2(\mathbb{R}) \rightarrow P_2(\mathbb{R})$ and is a linear endomorphism given by $T(f(x))=f(x)-f(2x-1)$.

Then I have to write$V=P_2(\mathbb{R})$ as a direct sum of $V=W_1\oplus W_2 \oplus W_3$ of T-invariant subspaces of dimension at least one.

So I now that $P_2(\mathbb{R}) = \{1,x,x^2\}$, I'm just confused on how to find the direct sum of it.

I know that in $V=W_1\oplus W_2 \oplus W_3$ all of the $W_i$ are T-invariant so this would translate to $V=K(T-\lambda _1 I)^{m_1}\oplus K(T-\lambda _2 I)^{m_2}\oplus K(T-\lambda _3 I)^{m_3}$ and they are all T-invariant. So I'm just confused, is that the answer or do I need to find of basis for T with respect to a basis for $P_2(\mathbb{R}) = \{1,x,x^2\}$?

• You have to compute the $\lambda_i$ in this concrete example and find the corresponding eigenspaces. Those will be the $W_i$. Clearly, to do this you have to write down explicitely the effect of $T$ on a basis as first step. – AdLibitum Mar 18 '15 at 23:37

First a comment to the solution one should expect: Since all of $W_i$ should have dimension at least one and $P_2(\mathbb{R})$ is $3$-dimensional, the dimension of each $W_i$ will be one.
So, what does $T$-invariant mean for a one-dimensional vector space? Let $W_i=\langle w_i\rangle$. Then this is $T$-invariant if and only if there exists a $\lambda_i$ with $T(w_i)=\lambda_i w_i$. Hence, what we are searching for are eigenvectors for $T$.
I will now just sketch what to do to attack this standard linear algebra problem. First, it is good to compute the matrix of $T$ with respect to a particularly easy basis. In your question you say that you already know one, namely $\{1,x,x^2\}$. We get that $T(1)=0$, $T(x)=-x+1$, $T(x^2)=-3x^2+4x-1$. Thus, the matrix of $T$ with respect to this basis is given by $$A_T=\begin{pmatrix}0&1&-1\\0&-1&4\\0&0&-3\end{pmatrix}.$$ Its characteristic polynomial is $\lambda(\lambda+1)(\lambda+3)$, hence the eigenvalues are $\lambda_1=0, \lambda_2=-1, \lambda_3=-3$. Notice that these are the same $\lambda_i$ that we wanted to construct before. To decompose $P_2(\mathbb{R})$ we need to find the corresponding eigenvectors (which in this case are polynomials, since $P_2(\mathbb{R})$ has polynomials as vectors). It turns out that $w_1=1, w_2=-x+1, w_3=x^2-2x+1$.