Given two square matrices $A, B$, when is $$\det(A+tB) = 0$$ for all $t\in \mathbb{R}$?  An easy sufficient condition is that $A$ and $B$'s kernels have nontrivial intersection. Per Henning's comment below, this is not also necessary.  Does there exist a nice necessary and sufficient characterization?
121890
I'd like to know if the following problem of elementary linear algebra is already solved / solvable.  For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?
116545
Find two $2 \times 2$ matrices $A$ and $B$ with the same rank, determinant, trace and characteristic polynomial, but that are not similar to each other.   I come up with two matrices: $A=\begin{pmatrix} 0 &amp;1 \\  0 &amp;0  \end{pmatrix}$ and $B=\begin{pmatrix} 0 &amp;0 \\  1 &amp;0  \end{pmatrix}.$ It is easy to check that they have same rank, determinant, trace and characteristic polynomial. However, my question is I do not know how to prove two matrices are similar or not. I have learnt the converse in my textbook, i.e. If two matrices are similar, they have the same determinant, characteristic polynomial,etc. I have also known that (but I do now know the proof), we can check by using Jordan form of two matrices. I do not know if this claim is correct: &quot;If two matrices have the same Jordan form, they are similar to each other.&quot; Yet, go back to the question, is it a quick way to prove? Thank you in advance.
52087
I'm trying to prove the following: Let $A$ be a  $k\times k$ matrix, let $D$ have size $n\times n$, and $C$ have size $n\times k$. Then,  $$\det\left(\begin{array}{cc} A&amp;0\\ C&amp;D \end{array}\right) = \det(A)\det(D).$$  Can I just say that $AD - 0C = AD$, and I'm done?
61531
How can you show that $$ \det(A + B^T) = \det(A^T + B)$$ for any $n\times n$ matrices $A$ and $B.$
84739
The determinant of a square matrix is zero if and only if the column vectors are linearly dependent.   I see a lot of references to this all over the web, but I can't find an actual explanation for this anywhere.
26113
Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.  If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.  Notation: $0 \le A$ means that $A$ is positive.  If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$. How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants.
11965
struggling with a question from homework and would appreciate some assistance.  Let $A, B \in M_2^{\mathbb{R}}$ be defined as follows: $$A = \begin{pmatrix} 2 &amp; 1 \\ 1 &amp; 1 \end{pmatrix}, B = \begin{pmatrix} -4 &amp; 1 \\ 1 &amp; 1 \end{pmatrix} $$ Find a regular $P \in M_2^\mathbb{R}$ such that $P^tAP=I$ and $P^tBP$ is diagonal.  I'm familiar with the theorem that states that a simultaneous diagonlization exists if one of the matrices is positive-definite (or negative-definite). I've found an invertible P such that $P^tAP=I$: $$P=\begin{pmatrix}1\over\sqrt{2}&amp;-1\over\sqrt{2}\\0&amp;\sqrt{2}\end{pmatrix}$$ But $P^tBP$ is not diagonal. Would appreciate any assistance / hints. Thanks!
78573
Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?  If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?
21095
I was hoping that someone could point me in the right direction on how I could start answering the following question:     Based on the fact that $\det(AB)=\det(A)\det(B)$ for all square $n \times n$ matrices $A,B$, explain in every detail why:   $$\det(A^{-1}) = 1/\det(A).$$
83834
The  mentions the following matrix identity $$\det\begin{pmatrix}A&amp;B\\B&amp;A\end{pmatrix}=\det(A+B)\det(A-B),$$ valid for squared matrices $A$ and $B$ of the same size.  How is this result proved?
71598
We have to find $\det(A^{2}+A^{T})$. It is given that eigenvalues of $A$ are $1,2,3$.  My attempt: Since the question implicitly states that the answer would be same for all $A$ with eigenvalues $1,2,3$ we find that the answer is $(1+1)*(2+4)*(3+9)$ by taking $A$ as a diagonal matrix :P  I am unable to prove it for general $A$. Any help will be appreciated.
18536
Prove that if $A$ is a square matrix and $\mathrm{Tr}(A)=0$, then there exists an invertible matrix $R$ such that the matrix $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$.   It seems like the formula $A=S\Lambda S^{-1}$, but maybe it does not help.  Thanks so much.
75546
Assume that the matrices $A,\: B\in \mathbb{R}^{n\times n}$ satisfy  $$ A^k=0,\,\, \text{for some $\,k\in \mathbb{Z^+}$}\quad\text{and}\quad AB=BA. $$  Prove that $$\det(A+B)=\det B.$$
121710
As in the title:  If two real (square) matrices are conjugated over $\mathbb{C}$, are they then also conjugated over $\mathbb{R}$?
16874
Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?   I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?
57195
Suppose that each of the row sums of an $n\times n$ matrix $A$ is equal to zero. Show that $A$ must be singular.  I can show this for $2\times2$ by using the fact that $det(A)=0$, whenever $A$ is singular. But do not know how to make it valid of for all square matrices.  Thanks!
41069
For injective side, in my opinion, we can find two square matrices $A, B$ with each injective but $AB$ has a zero row ($\dim\ker(AB)\ne 0$), so $TS$ is not injective. But I don't know how to find such matrices.  For surjective side, I have no idea how to start.  I feel my idea is very confusing, could someone give me a clear way to solve the problem?