Possible diagonal representation and the corresponding basis for $\mathbb R^3$

diagonalizationeigenvalues-eigenvectorslinear algebramatricesoperator-theory

Linear operator $T:\mathbb R^3\to\mathbb R^3$ is given by:
$$T(x_1,x_2,x_3)=(bx_1,bx_1+3x_2-x_3,-2x_1+2x_2)$$

Depending on the parameter $b\in\mathbb R$, determine if there exists a basis for $\mathbb R^3$ s.t. $T$ can be represented by a diagonal matrix. If exists, find such matrix representation and the corresponding basis.

My attempt:

For the purpose of easier computation of the eigenvalues, let $e=\{e_1,e_2,e_3\}$ be the standard base for $\mathbb R^3$ and let $f=\{Te_1,Te_2,Te_3\}$.

One case: $\forall b\ne 0\;f$ is another basis for $\mathbb R$.

Then $[T]_e^f=\begin{bmatrix}b&0&0\\b&3&-1\\-2&2&0\end{bmatrix}$.

$(*)$ Note: I have seen both notations $[T]_e^f$ and $[T]_f^e$, but I used the first one because it's used more often in our script.

Let $k_T=\det(T-\lambda I)=\begin{vmatrix}b-\lambda&0&0\\b&3-\lambda&-1\\-2&2&-\lambda\end{vmatrix}=(b-\lambda)(\lambda^2-3\lambda+2)=-(\lambda-b)(\lambda-2)(\lambda-1)$
$\implies \sigma(T)=\{1,2,b\}$

So, the algebraic and geometric multiplicities are equal.
$T$ could be represented by a diagonal matrix $A\in M_3$ with eigenvalues as diagonal entries:

$$A=\begin{bmatrix}b&0&0\\0&2&0\\0&0&1\end{bmatrix}$$

The wanted base consists of eigenvectors corresponding to different eigenvalues (because those are linearly independent). Let $B\in M_3$ be a transformation matrix whose columns are the eigenvectors. Then:
$$A=B^{-1}TB$$

I wanted to change the pair of bases $e,f$ into a single basis $f$, so
$$[T]_f^f=[T]_e^f[I]_f^e,\;\&\;\color{blue}{[I]_f^e=\left([I]_e^f\right)^{-1}},$$
but the choice of the basis $e$ yields the identity matrix.

I tried to solve the homogeneous systems $(T-\lambda I)x=0$ for each of the eigenvalues, and choose the eigenvectors from infinitely many non-trivial solutions, $b-\lambda x_1$ could imply $x_1$ always equals $0$. Then $B$ isn't invertible.
In the second case, $b=\lambda=0\implies\; T$ is singular.

My calculations got suspicious at this point and confused me.
May I ask for advice on how to solve this task? Thank you in advance!

I have also seen a related post: Diagonalizing the matrix (if possible).

Best Answer

After calculating the eigenvalues you cannot immediately say that the algebraic and geometric multiplicities are equal. We distinguish several cases.

  • $b \neq 1, 2$. Then $T$ has three distinct eigenvalues and is therefore diagonalizable. You can calculate three corresponding eigenvectors $v_1, v_2, v_3$ by solving $(T - \lambda_i E) v_i = 0$ for $i = 1, 2, 3$ and $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = b$. As a transformation matrix you can choose $B = (v_1\ v_2 \ v_3)$.

  • $b = 1$. Then the eigenvalue $\lambda = 1$ has geometric multiplicity $2$. But calculating the corresponding eigenspace $E_1$ yields $$E_1 = \operatorname{span}\{(0, 1, 2)^\intercal \}.$$ Since this space is one-dimensional, $T$ is not diagonalizable for $b = 1$.

  • $b = 2$. Similar to the second case. $T$ is not diagonalizable for

$$E_2 = \operatorname{span}\{(0, 1, 1)^\intercal \}.$$

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