1. ## Matrix Questions!

Ok, I've got 3 questions.

1) Given C and D are square matrices of order n, and inverse of C ( $C^{-1}$) exists, expand and simplify $(C^{-1} + CD)^2$ as much as possible.

My answer turns out to be

$(C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}$

Is this the most you can simplify it? And is the way I stated $(C^{-1})^2$ alright? Or should I do it like $C^{-1}C^{-1}$

2) Let u be any unit vector in $IR^n$, and let $A = I -$ uu $^T$ (where the T indicates that the matrix is transposed).

Prove that

a) $A = A^T$
b) $A^2 = A$

I can do a) but I cannot do b). Can anyone help me?

3)
a) For B a square matrix, expand the product (I + B)(I − B + B^2)
b) Let $B$ = $\begin{array}{ccc}0&p&q\\0&0&r\\0&0&0\end{array}$, and find $B^2$ and $B^3$
c) Using the matrix B and the result from part (b), simplify the formula in (a).
d) Let $A$ = $\begin{array}{ccc}1&p&q\\0&1&r\\0&0&1\end{array}$ = $B + I$ for B in b).

Use the results in (c) and (a) to find $A^{-1}$ .
Note: $I$ represents the identity matrix. Also recall that if $AC = CA = I$, if and only if , $C = A^{-1}$

Ok, for this question I can do all the way up to c but then I am lost how to do d). I mean I can do it by row reducing A and then finding the inverse that way. But the question says to use my results from a and c so I am thinking there must be abother way to do it. Can anyone help me out here?

My result for a) turns out to be I + B^3. Subsequently I find that B^3 is a 3x3 matrix with all zeros and thus my answer in c turns out to be just "I", the identity matrix.

2. Originally Posted by Johnaloa
Ok, I've got 3 questions.

1) Given C and D are square matrices of order n, and inverse of C ( $C^{-1}$) exists, expand and simplify $(C^{-1} + CD)^2$ as much as possible.

My answer turns out to be

$(C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}$

Check that last term. It's $CDCD$.

Looks to me like that's all you can do.

-Dan[/quote]

3. Originally Posted by Johnaloa
Ok, I've got 3 questions.

1) Given C and D are square matrices of order n, and inverse of C ( $C^{-1}$) exists, expand and simplify $(C^{-1} + CD)^2$ as much as possible.

My answer turns out to be

$(C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}$

Is this the most you can simplify it? And is the way I stated $(C^{-1})^2$ alright? Or should I do it like $C^{-1}C^{-1}$ As topsquark says, C^{2}D^{2} should be CDCD. Also, you can write C^{-2} for C^{-1})^2.

2) Let u be any unit vector in $IR^n$, and let $A = I -$ uu $^T$ (where the T indicates that the matrix is transposed).

Prove that

a) $A = A^T$
b) $A^2 = A$

I can do a) but I cannot do b). Can anyone help me? See below.

3)
a) For B a square matrix, expand the product (I + B)(I − B + B^2) I + B^3, correct.
b) Let $B$ = $\begin{array}{ccc}0&p&q\\0&0&r\\0&0&0\end{array}$, and find $B^2$ and $B^3$ B^3 = 0, correct.
c) Using the matrix B and the result from part (b), simplify the formula in (a). It becomes (I + B)(I − B + B^2) = I. You're still correct up to here.
d) Let $A$ = $\begin{array}{ccc}1&p&q\\0&1&r\\0&0&1\end{array}$ = $B + I$ for B in b).

Use the results in (c) and (a) to find $A^{-1}$. From c), I − B + B^2 is the inverse of I + B.
Note: $I$ represents the identity matrix. Also recall that if $AC = CA = I$, if and only if , $C = A^{-1}$

Ok, for this question I can do all the way up to c but then I am lost how to do d). I mean I can do it by row reducing A and then finding the inverse that way. But the question says to use my results from a and c so I am thinking there must be another way to do it. Can anyone help me out here?

My result for a) turns out to be I + B^3. Subsequently I find that B^3 is a 3x3 matrix with all zeros and thus my answer in c turns out to be just "I", the identity matrix.
For 2), let $P = \mathbf{uu}^{\textsc{t}}$. Then $P^2 = \mathbf{uu}^{\textsc{t}}\mathbf{uu}^{\textsc{t}} = \mathbf{uu}^{\textsc{t}} = P$ (because u is a unit vector and so $\mathbf{u}^{\textsc{t}}\mathbf{u}=1$). Therefore $A^2 = (I-P)^2 = \ldots$ (you do the rest).