$T$ is an endomorphism from $\mathbb R^3$ to $\mathbb R^3$

The eigenvalues of $T$ are $ \lambda\ _1=1 , \lambda\ _2=-1$

$G= \{(x,y,z)\mid x-y+z=0\} $ is an eigenspace of $T$


I need to prove $T$ is diagonalizable

In my attempt I found a basis of $G$, which is $\{(1,0,-1),(0,1,1)\}$

Then note $T(1,1,1)=(-1,-1,-1)$ have the eigenvalue $-1$

Can I assume the diagonal matrix have the eigenvalues $ \lambda_1=1 , \lambda_2=-1, \lambda_3=-1$ because (1,1,1) does not belong to the subspace spanned by $(1,0,-1),(0,1,1)$?


The first eigenspace has dimension $2$, the second dimension $1$. The sum is indeed $3$, the dimension of $V=\mathbb R^3$.

Thus, $T$ meets the requirement to be diagonalizable.

The diagonal matrix will in fact have two $1$'s and one $-1$ on the diagonal...


The point is that $v_3=(1,1,1)^T$ is not in the eigenspace $G=span(v_1,v_2)$, so they together span the 3d space.

Since $Tv_3=-v_3$, $\ v_3$ belongs to the eigenspace for $-1$, and $1$ is also an eigenvalue, so it goes for $G$.


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