One derivation I was verifying holds true if:  $$Q^{-1}\left(Q^{-1}+P^{-1}\right)^{-1}P^{-1}=\left(Q+P\right)^{-1}$$  where Q and P are positive definite matrices.  I've done some numerical tests and I incline to believe that.  But, where does this come from? I could not derive it from .
123189
I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$  I'm a bit lost with this one, I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$, Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.  If someone could clue me in I'll be grateful.    UPDATE:  Thanks for all of your input - it really helped, I got a solution, I'd like to know if my way of proving it is valid:  $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$      multiplying both sides by $A^{-1}$ and $B^{-1}$    $A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$    now multiplying both sides by $(A + B)$, I get to  $(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$  Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$    Now I determine that $C$ is $(A+B)^{-1}$, because $(A+B)C$ equals the Identity Matrix.    So lastly to verify that I place $C$ in the original equation:  $(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$  and from this I get:  $(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$
28988
Let $A$ and $B$ be real square matrices of the same size. Is it true that $$\det(A^2+B^2)\geq0\,?$$  If $AB=BA$ then the answer is positive: $$\det(A^2+B^2)=\det(A+iB)\det(A-iB)=\det(A+iB)\overline{\det(A+iB)}\geq0.$$
15449
$A$ and $B$ are $n\times n$ matrices, A = $B^{T}B$. Prove that if $\operatorname{rank}(B)=n$, then  $A$ is positive definite, and if $\operatorname{rank}(B)&lt;n$, $A$ is positive semi-definite.    My current understanding is that if $\operatorname{rank}(B)=n$, then $\operatorname{rank}(B^{T}B)=n$ then $\operatorname{rank}(A)=n$, making $A$ positive definite.   Similarly, if $\operatorname{rank}(B)\le n$, then $\operatorname{rank}(B^{T}B)\le n$ then $\operatorname{rank}(A)\le n$, making $A$ positive semi-definite.  This doesn't feel like a complete proof. Is there anything I am missing?
102331
show that determinant $$\left|\matrix{ x^2+L  &amp;  xy   &amp;   xz \\ xy   &amp;    y^2+L &amp;   yz \\ xz   &amp;    yz   &amp;   z^2+L \\ }\right| =  L^2(x^2+y^2+z^2+L)$$  without expanding by using the appropriate properties of determinant.  All i can do is LHS  $$x^2y^2z^2\left|\matrix{ 1+L/x^2  &amp; 1  &amp;   1 \\ 1   &amp;   1+L/y^2 &amp;   1 \\ 1  &amp;    1   &amp;   1+L/z^2 \\ }\right|$$
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 $Q$?
57684
I got a positive definite matrix $B,$ that is, $V(x) = x^TBx &gt; 0$ for any vector $ x \neq 0.$  I want to show that $ \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2$ for any $x \neq 0,$ where $\lambda_\min$ and $ \lambda_\max$ are defined by $$ \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ and  $$ \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B  \rbrace$$  Any hint please?
92343
As far as I know $(AB)^{-1}=B^{-1}A^{-1}$ and matrix multiplication is associative so:  $$(X′ X)^{−1}X′=(X^{-1})(X')^{-1}(X')=X^{-1}.$$  What am I doing wrong?
63496
P= $\left(\begin {matrix}q&amp;p&amp;0&amp;0\\0&amp;q&amp;p&amp;0\\0&amp;0&amp;q&amp;p\\0&amp;0&amp;0&amp;1\end{matrix}\right)$  We are to find $P^n$  using method of generating function.  I have understood the steps up to finding the inverse as inv (I-sP) = $\frac{adj(I-sP)}{|I-sP|}$  This matrix works out to   P= $\left(\begin {matrix}(1-sq)^{-1}&amp;sp(q-sq)^{-2}&amp;s^2p^2(1-sq)^{-3}&amp;s^3p^3(1-s)^{-1}(1-sq)^{-3}\\0&amp;(1-sq)^{-1}&amp;sp(1-sq)^2&amp;s^2p^2(1-s)^{-1}(1-sq)^{-2}\\0&amp;0&amp;(1-sq)^{-1}&amp;sp(1-s)^{-1}(1-sq)^{-1}\\0&amp;0&amp;0&amp;(1-s)^{-1}\end{matrix}\right)$  I have not understood the next step in the textbook which says now collect coefficients of $s^n$ by expanding each element of $(I-sP)^{-1}$ in powers of s to get the matrix. The first row of this matrix is  $\left(\begin{matrix} q^n&amp;&amp; npq^{n-1}&amp;&amp;\frac{n(n-1)p^2q^{n-2}}{2}&amp;&amp;p^3(1+3q+6q^2+.......)\end{matrix}\right)$  Request guide as to how is this done for the first row of the matrix. I will work out the rest based on this
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
1959
If I have a positive semidefinite matrix $A_{n\times n}$, I want to transform it to positive definite matrix. I plug the following operation to the original matrix: $$B=A+\lambda I_{n}$$ where $\lambda &gt; 0$. I want to know whether B is positive definite matrix. Could anyone proof it?
15155
So my friend and I are working on this and here is what we have so far.  We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?  Thank you.
36465
I can't understand why $\left(\frac{\partial P}{\partial V} \right)_T=-\left(\frac{\partial P}{\partial T} \right)_V\left(\frac{\partial T}{\partial V}\right)_P$. Why does the negative sign arise? I can easily write $\left(\frac{\partial P}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_P$ from the rule of partial derivative, but what's the negative sign for?
19615
How prove the following equality for a block matrix? $$\det\left[\begin{array}[cc]\\A&amp;C\\ 0&amp;B\end{array}\right]=\det(A)\det(B)$$ I tried to use a proof by induction but I'm stuck. Is there a simpler method? Thanks for help.
1461
First, my apologies if this has already been asked/answered.  I wasn't able to find this question via search.  My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2.  In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof.  As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.  If we write equations (3) as $q = p - (p^2 - 2)x$, we can write (4) as   $$ q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. $$  Here, we need a rational $x &gt; 0$, chosen such that the expression in $[...]$ is positive.  Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need   $$ x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, $$  or, for $p \in B$, $x &lt; 1/\left(p + \sqrt{2}\right)$ or $x &gt; 1/\left(p - \sqrt{2}\right)$.  Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$.  Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.  The Question:  Why does Rudin choose $x = 1/(p + 2)$ specifically?  Is it just to make the expressions work out clearly algebraically?  Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof?  Is there a simpler derivation for the number $q$ that I am missing?
72842
How do I derive:   $\ln(-1)=i\pi$ and   $\ln(-x)=\ln(x)+i\pi$  for $x&gt;0$ and  $x \in\mathbb R$  Thanks for any and all help!
78948
I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$
21585
I need help proving or disproving this! If anyone could show me the sequence of steps I can use to help me show the left is equal to the right I would be grateful.  $$\mathcal P(A-B)=\mathcal P(A)-\mathcal P(B)$$