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Let $V$ be a finite dimensional vector space over $\Bbb R$ and let $T :V \to V$ be a linear transformation. Let $A$ be the matrix of $T$ with respect to the standard basis for $V$. For each of the following assertions, state whether it is true or not. If the assertion is true, briefly justify why this is so; otherwise, provide a counter-example.

(a) If $T$ is diagonalisable then each of its eigenvalues has algebraic multiplicity equal to $1$.

(b) If none of the eigenvalues of $T$ are zero then the determinant of $A$ is not zero.

(c) If $v_1$ and $v_2$ are eigenvectors for $T$ associated to eigenvalues $\lambda_1$ and $\lambda_2$, respectively, then $v_1 + v_2$ is an eigenvector for $T$ with associated eigenvalue $\lambda_1 + \lambda_2$.

For (a) I gave a counterexample and said it was false. Is this the correct answer?

For (b) I think it is correct since the determinant of characteristic polynomial of $A$ can only be zero if there is no eigenvalue. Is this correct? If so is there a theorem that proves that?

For (c) I said it is false since eigenvalues added together does not form a new eigenvalue for $T$, is this correct?

Thanks for the help!

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  • $\begingroup$ Each of your truth values is correct. For $(b)$, there is a theorem which says that the determinant is the product of the eigenvalues. So $\det(A) = 0 \iff$ at least one of the eigenvalues is $0$. For $(c)$, remember that you're supposed to come up with a counterexample. There are simple ones. $\endgroup$ – user137731 Nov 1 '16 at 21:26
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Assertion (a) is false, as you say. $T$ the identity map is a counterexample. Assertion (b) is true, since the determinant of $T$ is the product of its eigenvalues (with multiplicity). Assertion (c) is false, as we can see when $T$ is scalar multiplication by 2. Every eigenvalue is 2; 4 is not an eigenvalue.

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