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 $P_2(\mathbb{R}) = \{1,x,x^2\}$?

  • $\begingroup$ 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. $\endgroup$ – 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 $w_1=1, w_2=-x+1, w_3=x^2-2x+1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.