# Tag Info

8

Let $\lambda \in \mathbb C$ be an eigenvalue with eigenvector $0 \neq x \in \mathbb C^n$. Then $$(x,x) = (A^tAx,x) = (Ax,Ax) = \lvert\lambda \rvert^2 (x,x).$$ Thus $\lvert \lambda \rvert = 1$ for all eigenvalues. Since $A$ is real, eigenvalues come in conjugate pairs, so either all three are real, or two are complex and one is real, but the two which are ...

4

Suppose $\lambda$ is an eigenvalue of $A$ with associated eigenvector $x\neq0$. Then we have $$|\lambda|\cdot|x|=|\lambda x|=|Ax|>|x|,$$ and thus $$|\lambda|>1.$$ Since the eigenvalue $\lambda$ was arbitrary, we can conclude that $|\lambda|>1$ for all eigenvalues of $A$

4

What about $\textbf{A}:=\begin{bmatrix}8&12\\-3&-4\end{bmatrix}$? This matrix has only one eigenvalue $2$, which is greater than $1$. However, under the standard norm on $\mathbb{R}^2$, the size of $\textbf{A}\,\textbf{u}$ is less than the size of $\textbf{u}$, if $\textbf{u}:=\begin{bmatrix}3\\-2\end{bmatrix}$. You can even require that the ...

4

I presume the statement is that $A$ is a 2 by 2 matrix with $p(\lambda)=\det(\lambda I_2 -A) = \lambda^2+4 \lambda -12$? The roots ($-6$ and $2$) are the eigenvalues. Eigenvalues behave nice when taking powers so if $\lambda$ is an eigenvalue of $A$ then $\lambda^3$ is an eigenvalue of $A^3$, etc... Do you need a proof? The last part is less trivial, I ...

3

Suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. If you apply the given equation to the vector $v$, you find \begin{align} 0 &= (A^2 - 3A + 2I)v \\ &= A^2v - 3Av + 2v \\ &= A(\lambda v) - 3\lambda v + 2v \\ &= \lambda^2v - 3\lambda v + 2v \\ &= (\lambda^2 - 3\lambda + 2)v \end{align} So what does this tell you ...

2

Assuming $A$ is a real matrix, using singular value decomposition we can write $$A = U S V^T$$ where $S$ is a real valued diagonal matrix (i.e., $S=S^T$); $U$ is the left Eigenvector and $V$ the right Eigenvector. Then, you can write $$A^TA = V S^T U^T U S V^T = VS^2V^T$$. However, if $A$ is positive symmetric, then $U=V$ and you can use eigenvalue ...

2

Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.

1

Hints: For a), it is useful to note that $A = xx^T$, where $x$ is the column-vector $x = (a,b,c)$. No, the number of minors that are zero is not necessarily the multiplicity of the zero eigenvalue; if you wanted to use minors, you'd have to show that every $2 \times 2$ submatrix has determinant zero. You should instead find the rank of the matrix through ...

1

a) I agree. I am not sure about your conjecture how the minors relate to the multiplicity of $\lambda=0$ However, I can eyeball two eigenvectors for $\lambda=0$ $v_1 = \begin{bmatrix} b\\-a\\0\end{bmatrix}, v_1 = \begin{bmatrix} c\\0\\-a\end{bmatrix}$ b) looks fine c) For each $X$, Suppose $P,Q \in X$ then for example iii) if $trace (AP)=0$ and $trace (... 1 If it's an$n \times n$matrix with all diagonal entries$1$, the trace is$n$, so that won't help. The Frobenius norm is bounded by$n$. Since the smallest eigenvalue can be arbitrarily close to$0$, I don't see how you could possibly get a nonzero lower bound in terms of the Frobenius norm. 1 By$λ^2 + 4λ - 12 = 0$,$λ = 2 $or$λ = -6 $Through diagonalization, A is similar to D= \begin{bmatrix} 2 & 0\\ 0 & -6 \end{bmatrix}$A^3$is similar to$D^3$= \begin{bmatrix} 8 & 0\\ 0 & -216 \end{bmatrix} So eigenvalue for$A^3$is$8 $and$-216$. 1 There is no way to "see" the eigenvectors directly, no more than it is easy to predict the solution of any given linear system of equations. Because this is exactly what the problem is; solving $$(A - \lambda I)x = 0.$$ Of course, in some cases, it can be done, e.g. if$A - \lambda I$turns out to be a matrix with some special structure. But to a general ... 1 It's because$-A^2 = A^T A$has only real nonnegative eigenvalues: if$v$is an eigenvector of$A$with eigenvalue$\lambda$then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$ Skew-symmetric matrices do not have to have nonzero determinant, the zero matrix is a counterexample. 1 Exactly like in the link you did. The upper left hand 2 by 2 block will be the one for the first pair, the lower right hand 2 by two block will be the one for the second pair, and the upper right and lower left will be all 0s. 1 $$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \lambda_{\min} (\mathrm A) > 0$$ because$\|\mathrm x\|_2 = 1$and$\mathrm A \succ \mathrm O$. The minimum is attained at the intersection of the eigenspace of$\lambda_{\min} (\mathrm A)\$ with the unit Euclidean sphere.

1

T(B)=0 i.e AB=0,but A is invertible.This implies B=0. Hence kerT={0}. T is one-one.T is Onto also.Hence T is invertible.

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