Page 430 - 35Linear Algebra

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T
because if P = [v 1 , v 2 , v 3 ] is created from orthonormal vectors then P −1 = P ,
which means computing P −1 should be easy. So lets say

√ √ 0
 1   1   
2 2
1
v 1 =  − √  , v 2 =  1  , and v 3 =  0 
√
2 2
0 0 1
so we get
√ √ 0 √ − √ 0
 1 1   1 1 
2 2 2 2
√
P =  − √ 1 √ 1 0  and P −1 =  1 √ 1 0 
2 2 2 2
0 0 1 0 0 1
So when we compute D = P −1 MP we’ll get
√ − √ 0 1 2 0 √ √ 0 −1 0 0
 1 1     1 1   
2 2 2 2
 1 √ 1 0   2 5 − √ 1 √ 1 0  =  0 3
√
2 2 0   2 2 0 
0 0 1 0 0 5 0 0 1 0 0 5
Hint for Review Problem 1
2
For part (a), we can consider any complex number z as being a vector in R where

1 0
complex conjugation corresponds to the matrix . Can you describe z¯z
0 −1
in terms of kzk? For part (b), think about what values a ∈ R can take if
a = −a? Part (c), just compute it and look back at part (a).
†
For part (d), note that x x is just a number, so we can divide by it.
Parts (e) and (f) follow right from definitions. For part (g), first notice
that every row vector is the (unique) transpose of a column vector, and also
T T
think about why (AA ) = AA T for any matrix A. Additionally you should see
†
T
that x = x and mention this. Finally for part (h), show that
†
†
x Mx x Mx T
=
†
†
x x x x
and reduce each side separately to get λ = λ.
G.15 Kernel, Range, Nullity, Rank

Invertibility Conditions
Here I am going to discuss some of the conditions on the invertibility of a
matrix stated in Theorem 16.3.1. Condition 1 states that X = M −1 V uniquely,
which is clearly equivalent to 4. Similarly, every square matrix M uniquely

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