Proving that a linear map is diagonalizable and finding a similar diagonal matrix

linear algebramatrices

Given the linear map $T: \Bbb R_3[X] \to \Bbb R_3[X]$ that is defined as
$T(ax^2+bx+c)=(a+b+c) \cdot x^2 +(2a+2b+2c) \cdot x +a+b-c$ prove that $T$ is diagonalizable and find a similar diagonal matrix without calculating the diagonalizable matrix.

I am still very new to this topic so I tried using the basic definitions and way of solving.

First thing was to find Eigenvalues . so we need $\lambda \in \Bbb R$ such that for $0 \not= v \in \Bbb R_3[X]$ we get $T(v)= \lambda \cdot v$ so we need to look for $v=ax^2+bx+c \in \Bbb R_3[X]$ such that

  1. $T(ax^2+bx+c)= \lambda \cdot (ax^2+bx+c)$

  2. $T(ax^2+bx+c)=(a+b+c) \cdot x^2 +(2a+2b+2c) \cdot x +a+b-c$

so we have 2 equations , we compare and get $(a+b+c) \cdot x^2 +(2a+2b+2c) \cdot x +a+b-c = \lambda \cdot (ax^2+bx+c)$

we now have 3 equations:

  1. $a+b+c = \lambda \cdot a$ $\iff$ $a(1-\lambda)+b+c=0$

  2. $2(a+b+c)= \lambda \cdot b$ $\iff$ $2a +b\cdot(1-\lambda)-2c=0$

  3. $a+b-c= \lambda \cdot c$ $\iff$ $a+b+c(-1-\lambda)=0$

we need the solution to be different than zero so we will first put them in a matrix $ $$\left(\begin{matrix}
1-\lambda & 1 & 1 \\
2 & 2- \lambda & -2 \\
1 & 1 & -1-\lambda \\
\end{matrix} \right)
$ $ $

we will need a singular matrix so if we solve a determinant and get a solution equal to zero it means it has a non trivial solutions
$ $$\left|\begin{matrix}
1-\lambda & 1 & 1 \\
2 & 2- \lambda & -2 \\
1 & 1 & -1-\lambda \\
\end{matrix} \right|
$ $ $ $=-\lambda^3+2\lambda^2+2\lambda$ from here our Eigenvalues are $\lambda_1= 1+\sqrt3$ ,$\lambda_2= 1-\sqrt3$ and $\lambda_3= 0$
we have 3 Eigenvalues and our space dimension is 3 so the linear map is diagonalizable

Notes:

  1. In the book the final answer is different they got $\lambda_1= 1+\sqrt7$ ,$\lambda_2= 1-\sqrt7$ and $\lambda_3= 0$ but they do not show the way .. what did I do wrong? I solved according to the definitions

2.How can I find the diagonal matrix?

  1. $\Bbb R_3[X]$ is the polynomial that are with power less than 3

Thanks for any help and tips , and sorry if the translations are not correct hope it is understandable!

Best Answer

  1. It's due to a computational error. Your second equation should be $2a + (2-\lambda)b + 2c = 0.$ (And this yields the correct result, I've checked it.) Otherwise, your reasoning is totally sound! Although, in this stage of linear algebra, you should know that you can do things considerably shorter. Since $\mathbb{R}_3[X]$ behaves "basically the same" as $\mathbb{R}^3$, you can "translate" your problem into equivalent one in $\mathbb{R}^3$, solve it, and then "translate back" to give your answer in $\mathbb{R}_3[X]$ (you can see $\mathbb{R}^3$ as the space of the coordinates of your vectors in $\mathbb{R}_3[X]$ with respect to the standard $\mathbb{R}_3[X]$ basis).

  2. Without converting your problem into the equivalent $\mathbb{R}^3$ problem, you can't really talk about the diagonal matrix. If you'd do that, you'd find that the diagonal matrix is the matrix with your eigenvalues on the diagonal.

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