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9

As you said, $A$ is a square matrix. Since $A^2=0$, then $$0=\det 0=\det(A^2)=\det(A)^2$$ so $\det A=0$, and this means that the rows and columns of $A$ are LD.

4

Let $\lambda_{1,2,3}$ denote the eigenvalues of $A$, then $\operatorname{det}(A+I)=\operatorname{det}(A+2I)$ can be rewritten as $$(\lambda_1+1)(\lambda_2+1)(\lambda_3+1)=(\lambda_1+2)(\lambda_2+2)(\lambda_3+2),$$ which is in turn equivalent to (check it!) $$... 4 Yes this is correct as long as you assume that A is a 3 \times 3 matrix, except tr(A^2) = 1^2+ 2^2 + 4^2 = 21, since Av = \lambda v \implies A^2v = \lambda^2v 2 Multiplying a diagonal matrix by itself, A^2, will result in each of the diagonal entries being squared. For example, your matrix A has diagonal entries 1, 2, and 4. A^2 has entries 1, 4, and 16. The trace of this matrix is 21. 2 "Singular" means that 0 is a third eigenvalue. With three distinct eigenvalues 0,-3,-4, P must be diagonalizable. P^2 + 3P is also diagonalizable with respect to the same basis as P. 2 The triangular matrices are not normal in general. Only the diagonal matrices are. So take a conjugate of$$A=\begin{bmatrix} 0 & 1 & 1\\0 & 0 &1\\0 & 0 &0\end{bmatrix}$$For example, consider PAP^{-1} with$$P=\begin{bmatrix} 1 & 0 & 0\\0 & 1 &1\\1 & 0 &1\end{bmatrix}$$which happens to be ... 2 To construct examples of this, it's useful to take the 2\times 2 case and extend it to larger sizes First observe that there are no diagonalizable matrices with this property. Thus, such matrices, if they exist are similar to either \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix} or \begin{bmatrix}0 & 1 ... 2 I assume A is a square matrix. If the columns of A were linearly independent, then the image space of A would be the whole space and so A would be invertible. But this and A^2=0 imply A=0 (see \star below), which contradicts the columns of A being linearly independent, since the columns of the zero matrix are linearly dependent. (\star) ... 2 Since you have already shown that A has an eigen value as 0, therefore \exists \, x \neq 0 such that Ax=0. Thus by definition there exists a non-trivial linear combination of the columns of A which equals 0, hence columns are linearly dependent. 2 What an interesting question! I think the answer is yes. Consider the eigendecomposition of the self-adjoint matrix,$$A=Q D Q^*,$$where Q is unitary and D diagonal. The orthogonal projector onto an eigenspace is$$P_\lambda = Q I_\lambda Q^*,$$where I_\lambda is the diagonal matrix with ones in locations corresponding to the eigenspace of ... 1 Here's a positive definite counterexample:$$\begin{bmatrix}2&1&0\\1&2&0\\0&0&1\end{bmatrix}.$$1 For another perspective: If a column of A is zero, then the columns are clearly linearly dependent. Otherwise, each column of A lies in the kernel of the map v \mapsto Av, hence each column of A determines a linear dependence among the columns of A. 1 There's two possible interpretations here. Either you mean that this represents a linear map \mathbb{C} \to \mathbb{C} by z \mapsto e^{-i\theta}z in which case the problem is quite simple: without even invoking the algebraic completeness of \mathbb{C}, for any field F, a "linear map" T:F \to F, can be completely decscribed by the data T(1) = ... 1 If the geometric multiplicity of 0 is two which means that the dimension of the eigenspace of 0 is 2 then there's two linearly independent eigenvectors associated to 0 and then the given matrix would be similar to$$\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix} and this is a contradiction. Think to the rank of the two ...

1

This is (still) trivially false if you just take a small epsilon $p_{12}=p_{21}=1-\epsilon$ and $p_{11}=p_{22}=\epsilon$ The trace of $P^2$ is $2((1-\epsilon)^2+\epsilon^2)>2\epsilon$, if take $\epsilon$ to be, say, smaller than 0.2

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