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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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