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Diagonalization

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers

Find a basis B for the domain of T such that the matrix for T relative to B is diagonal.
T: R2 → R2: T(x, y) = (6x + 3y, 2x + y)
Not sure where to even start with this one.
I have given a link to the topic there so the OP can see my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
We can express $T \begin{pmatrix}{x}\\{y}\end{pmatrix}=A \begin{pmatrix}{x}\\{y}\end{pmatrix}$ with $A=\begin{pmatrix}{6}&{3}\\{2}&{1}\end{pmatrix}$. Eigenvalues of $T$: $$\det (A-\lambda I)=\lambda^2-(\mbox{trace }A)\lambda+\det A=\lambda^2-7\lambda=0\Leftrightarrow \lambda=7\vee \lambda=0$$ both real and simple, so $T$ is diagonalizable over $\mathbb{R}$. The eigenspaces are: $$\ker (A-7I)\equiv\left \{ \begin{matrix}-x+3y=0\\ 2x-6y=0\end{matrix}\right.\;,\quad \ker (A-0I)\equiv\left \{ \begin{matrix}6x+3y=0\\ 2x+y=0\end{matrix}\right.$$ with respective basis $B_{7}=\{(3,1)^t\}$ and $B_{0}=\{(-1,2)^t\}$. As a consequence, $B=\{(3,1)^t,(-1,2)^t\}$ is a basis of eigenvectors for $\mathbb{R}^2$ and the matrix for $T$ relative to $B$ is $D=\begin{pmatrix}{7}&{0}\\{0}&{0}\end{pmatrix}.$