Set of linear algebra questions Are the following statements true/false, explain:


*

*There is no such matrices $A$ and $B$ such that $C(A)=C(B)$ and $N(A)\neq N(B)$.

*If for a matrix $A\in M_{3x3}\left(\mathbb{R}\right)$ the following conditions hold true: $AA^T=A^TA=I$ and $\det A=1$ then $p_A(1)=0$.

*If $A\in M_{3x3}\left(\mathbb{R}\right)$ $\lambda _1=-1,\lambda \:_2=0,\lambda \:_3=1$ what is $\det(I+A^{100})$?

*If $A$ is a symmetric matrix and $\operatorname{tr}(A^2)=0$ what can you say about $A$?


i) $A$ is a zero matrix.
ii) $A$ is not a zero matrix.
iii) $0$ is the only eigenvalue of $A$.
iv) There is at least one non-zero eigenvalue for the matrix $A$.
My attempts:
The first statement is false, their kernels are of the same dimension, but they don't have to be spanned by the same vectors, can't think of a good example perhaps a matrix $A=(e_1,0)$ and $B=(0,e_1)$?
2) don't know where to start
the third one is tricky I know that $\det A=0$, but here it's not just $A$ by itself. Any tips?
4th well if it was $\operatorname{tr}(A)=0$, then $A$ would be a matrix with all zero eigenvalues, but is it different in this case?
EDIT: for the first problem if $C(A)=C(B)$ and $N(A)=N(B)$ are those two matrices then equal?
 A: *

*Your counterexample works. And no, even if $C(A) = C(B)$ and $N(A) = N(B)$ it doesn't mean that $A = B$. Take $A = (e_1,0)$ and $B = (2e_1,0)$.

*Your conditions say that $A$ is an orthogonal $3 \times 3$ matrix with determinant $1$. The question then asks whether $1$ is an eigenvalue of $A$. This is true and known as Euler's rotation theorem (you can find a linear algebra proof in the attached link).

*The matrix $A$ has three distinct eigenvalues and so it is diagonaliable. In other words, $A$ is similar to $\operatorname{diag}(-1,0,1)$. But then $I + A^{100}$ is similar to $I + \operatorname{diag}(-1,0,1) = \operatorname{diag}(0,1,2)$. Hence $\det(I + A^{100}) = \det(\operatorname{diag}(0,1,2)) = 0$.

*Since $A$ is symmetric, $A$ is diagonalizable with real eigenvalues $\lambda_1,\dots,\lambda_n$. But then $A^2$ is diagonalizable with eigenvalues $\lambda_1^2\dots,\lambda_n^2$ and if $\operatorname{tr}(A^2) = \sum_{i=1}^n \lambda_i^2 = 0$ then $\lambda_1 = \dots = \lambda_n = 0$ so $A = 0$.

