How to prove Binet theorem, namely $\det(AB)=\det(A)\det(B)$ ?  I've tried with the formulation of two arrays $2 \times 2$,  but could not.
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I've tried with the formulation of two arrays $2 \times 2$, but could not.
83671
Let $A$ be $m \times n$ matrix, and $B$ be $n \times m$ matrix. Then  Show that $$\det\begin{bmatrix}I_{n} &amp; B\\ A &amp; I_{m} \end{bmatrix}=\det\begin{bmatrix}I_{m} &amp; A\\ B &amp; I_{n}\end{bmatrix}$$  Show that $$\det(I_{m}-AB)=\det(I_{n}-BA)$$  Is it true that $I_{m}-AB$ and $I_{n}-BA$ have the same rank?    Someone please show me a way to start proving. I tried to use formulas for calculating block matrix with square matrices, but it didn't help me on this one. I really don't know where to start.
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Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
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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 $$\det(A+B)=\det B.$$
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Let $A,B\in M_{n}(\mathbb{R})$ be symmetric, with $A&gt;0$ and $B&gt;0$.  I need to prove that $\det (A+B)&gt;\max (\det(A), \det(B))$.  I want to use Sylvester theorem of having a matrix $D$ so that $D=\operatorname{diag}(1,1,\ldots, 1,-1,-1, \ldots-1,0,0,\ldots,0)$.  Do I need to use it? How do I use it here?  Thank you.
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Prove that if $A$ is a square matrix with integer entries and $\det(A) = \pm 1$, then $A^{-1}$ contains all integer entries.  I'm really thrown off by this one, its unlike all the examples I've seen.. I just dont know where to start... how can I begin to prove this?
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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?
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Let $A,\ B,\ C,\ D \in \mathcal{M}_n(\mathbb{C})$. If $\operatorname{rank}\left( \begin{bmatrix} A &amp;B \\ C &amp;D \end{bmatrix}\right)=n$, prove that $\det(AD)=\det(BC)$.
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Sylvester's determinant identity states that if $A$ and $B$ are matrices of sizes $m\times n$ and $n\times m$, then   $$ \det(I_m+AB) = \det(I_n+BA)$$   where $I_m$ and $I_n$ denote the $m \times m$ and $n \times n$ identity matrices, respectively.  Could you sketch a proof for me, or point to an accessible reference?
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I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!
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Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.  I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.  Thank you very much for your help!
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I find this question very interesting, but I am having trouble figuring out how to approach the problem.   I know that the $\det(A^{-1}) = 1/\det(A)$, but I'm unsure of where to go from here. If someone can send me in the right direction, it would be very much appreciated.   Thanks in advance,  Tommy.
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If $A^2=-I$ , where $A$ is a square matrix of order $n$ and which contains real entries only and $I$ is identity matrix. Then how can we prove that $\det(A)=1$?.  I could prove that $n$ should be an even integer. But could not proceed to prove that $\det(A)$ can take only $1$, finding out few matrices which satisfies such properies (of small order) also verifies the given statement that the determinant is only $1$ and not $-1$.  Can anyone help with a hint ?
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I need help with this proof:     Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.   My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.  Where do I go from here?
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Let A be an invertible $n \times n$ matrix whose inverse is itself. Prove that $\det(A)$ is either $1$ or $-1$.    I'm really lost in class. I don't even know where to start. Please help.
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Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.  I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction  n=2 we have the $A=\begin{bmatrix} 0 &amp; 1\\  1 &amp; 0 \end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1  Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$  Now to show for  a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
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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?