This problem comes from Linear Algebra and its Applications by David C Lay and Steven R Lay.  Let $c$ be a unit vector in $\mathbb{R}^3$. Consider the matrix $A = I - 2cc^T$   a ) What are the eigenvalues of $A$?   b) Find an easily computable eigenvector of $A$  My attempt: Firstly we can see that $cc^T$ is a symmetric $rank\ 1$ matrix. Let $v_i$ be a eigenvector of $A$ for $i = 1, 2, 3$ $$Av_i = \lambda_iv_i = (I - 2cc^T)v_i \\ = Iv_i - 2cc^Tv_i \\ = v_i  - 2cc^Tv_i$$ I'm stumped on where to go next. Is there something I'm missing? Any pointers would be greatly appreciated :)
123000
What would be the eigenvalues for a Householder matrix defined as: $H = I - 2 u u^T$? Can someone explain it  to me intuitively or with a simple proof?
75026
Let $A$ be a complex $3\times 3$ matrix such that $A^3=-I$, then we need to find out which of the following statements are correct?   $A$ has three distinct eigenvalues; $A$ is diagonalizable over $\mathbb{C}$; $A$ is triangulizable over $\mathbb{C}$; $A$ is non singular.   Wll, from minimal polynomial approach I have deduced that all are correct as minpoly has to be $x^3+1$ am I right?
77242
Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix ($A = -A^T$)  is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix.   Why does such a matrix have at least 2 non-zero eigenvalues?
12774
If the $\det(A)=0$ why does the matrix $A$ have an eigenvector? Explain why there is a basis $B$ in $R^n$ so that the matrix $[A]_B$ has the zero vector as its first column   I know that if the $\det(A)=0$ of a matrix is zero then it is singular and thus not invertible. That in turn means that it doesn't have one solution. Why does that mean that it has an eigenvector?  As for the second I have no idea.
62412
Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix. Prove that $\det(A) \geq 1$.  It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$ Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the  eigenvalues of $A$, so we get that $$(\lambda_1 - 2)(\lambda_2 - 2) \cdots (\lambda_n - 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$). Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
117274
I'm having troble proving this: Let $A\in M_n(R). $ If $n=2$ and $A$ has exactly one eigenvalue $\lambda$ then $A$ is similar to $\begin{bmatrix}  \lambda &amp; 1\\ 0 &amp; \lambda \\ \end{bmatrix}$ or to $\begin{bmatrix}  \lambda &amp; 0\\ 0 &amp; \lambda \\ \end{bmatrix}$ I did prove that if the geometric multiplicity of $\lambda$ is $2$, then $A$ is similar to $\begin{bmatrix}  \lambda &amp; 0\\ 0 &amp; \lambda \\ \end{bmatrix}$, but I'm struggling with the case when $A$ is not diagonalizable. Any thoughts?
12785
Let $A$ belongs to $M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases  when $A$ is diagonalizable over $\mathbb R$:  a. when $A^2 = A$; b. when $(A - 3I)^2 = 0$; c. when $A^2 + I = 0$  (a) is certainly true as eigen values are distinct. (c) is certainly not true as eigen values are imaginary not real. but i am not sure about (b). can anybody help? thanks.
40159
For any matrix $A$, $AA^T$ is symmetric, and thereby diagonalizable (Spectral Theorem). This implies that $AA^T$ has n linearly independent eigenvectors and therefore full rank.  Does this imply that for any matrix $A$, $AA^T$ is always invertible ? I know there is something wrong here because from the above chain we can conclude that a symmetric matrix is always invertible, which is definitely not true.  Could someone please point out where I am going wrong?
45709
We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks
76200
a.) Let A and B be $n$ x $n$ matrices. Prove that the matrix products   $AB$  and $BA$ have the same eigenvalues.      b.) Prove that every eigenvalue of a matrix A is also an eigenvalue of   its transpose $A^T$. Also, prove that if v is an eigenvector of A with   eigenvalue $\lambda$ and w is an eigenvector of $A^T$ with a different   eigenvalue $\mu \ne \lambda$, then v and w are orthogonal vectors with   respect to the dot product.   For a, I know that if their eigenvalues are the same then their eigenvectors must relate too.    For the first part of b, is it similar to proving that the $det(A) =det (A^T)$? And, I do not know how to do the second part.
63551
Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.   Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.   We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.  I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$  But I'm stuck when $v$ is not an eigenvector.  (Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates)
89209
Determine the number of possible values for $\det(A)$, given that $A$ is an $n \times n$ matrix with real entries such that $A^3 - A^2 -3A +2I=0$.   is the source of the problem. In the last comment, I don't understand why we can say for sure that $A$ has only three distinct eigenvalues. Can anyone explain to me?  Remark: I understand that the roots of the equation $x^3-x^2-3x+2=0$ are eigenvalues of $A$. My problem is why the matrix $A$ cannot have another eigenvalue?
112517
Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$.  Then all eigenvalues of $(A-iI)B$ are   purely imaginary real of modulus one of modulus less than one   My attempt: I discarded options $1,4$ by taking $A=O$ $($null matrix$)$. Let '$a$' be an eigenvalue of $A$ then the eigenvalues of $(A-iI)B$ are $(a-i)(a+i)^{-1}$. But I'm not sure about it. Please help me.
36892
Given a real, skew-symmetric matrix $\mathbf{A}\in\mathbb{R}^{m\times m}$, and a nonzero vector $X\in\mathbb{R}^{m}$, classify the scalar $\lambda$ (real, complex, imaginary, etc.) in the eigenvalue equation $$ \mathbf{A}X = \lambda X $$  The matrix $\mathbf{A}$ is skew symmetric if $\mathbf{A}^{T}=-\mathbf{A}$.  My try : I am new to matrices , so can not get a idea how to deal with it , Any hint will help
5906
There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$  I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle  Ax, x \rangle= \langle  x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero.  This argument doesn't work for a general complex skew-symmetric matrix.  Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case?  Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?  Thanks!
73571
I just wanted to make sure that the following statement is true:  Let $A$ be a normal matrix with eigenvalues $\lambda_1,...,\lambda_n$ and eigenvectors $v_1,...,v_n$. Then $A^*$ has the same eigenvectors with eigenvalues $\bar{\lambda_1},..,\bar{\lambda_n}$, correct?
9690
I would like to have a result similar to . Namely:      Let $R$ be a commutative $\mathbb{C}$-algebra, $A \in \mathcal{M}_n(R)$ such that $\mathrm{tr}(A)=\cdots=\mathrm{tr}(A^n)=0$. Does it follow that $A^n=0$?   I have no idea weather or not the analogy of eigenvalue exists for matrices over algebra so I can't simply generalize considerations from cited post. Any help will be appreciated.