# Math 2270 ```Math 2270
Quiz 6
1. Consider an n &times; n matrix A. Show that there are scalars c0 , c1 , . . . , cn , not all zero, such that the matrix
c0 In + c1 A + c2 A2 + &middot; &middot; &middot; + cn An
is noninvertible. [Hint: any n + 1 vectors from Rn will be linearly dependent.]
Let ~v ∈ Rn be any nonzero vector. Then the n + 1 vectors ~v , A~v , A2~v , . . . , An~v are linearly dependent. This means
that there exist constants c0 , c1 , . . . , cn such that
c0~v + c1 A~v + &middot; &middot; &middot; + cn An~v = ~0 ⇒ [c0 In + c1 A + &middot; &middot; &middot; + cn An ]~v = ~0.
This shows that ker([c0 I + c1 A + &middot; &middot; &middot; + cn An ]) 6= {~0}, which is the same as saying that this matrix is not
invertible.
2. Find the matrix B of the linear transformation T (~x) = A~x with respect to the basis B = {~v1 , ~v2 }, where
0 1
1
1
A=
, ~v1 =
, ~v2 =
.
1 0
1
−1
The formula given in class and in the text for the B-matrix of a linear transformation is
B = [[T (~v1 )]B &middot; &middot; &middot; [T (~vn )]B ],
where B = {~v1 , . . . , ~vn }. In our case, T (~vi ) = A~vi , so we compute
0 1
1
1
0 1
1
−1
A~v1 =
=
= ~v1 , A~v2 =
=
= −~v2 .
1 0
1
1
1 0
−1
1
The B coordinates are easily calculated in this case,
1
0
[A~v1 ]B =
, [A~v2 ]B =
.
0
−1
This gives us that
B=
1
0
0
−1
.
```