and M340L:
Matrices Matrix Calculations
Unique Number: 57160
Practice Final sketchy Solution, May , 2006
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13
1 1
I. Let A =
12
23
1) What is the domain
21
1
0 1 1
10
1
12
2
of A? What is
0
0
.
1
3
the co-domain of A? (3points)
The domain of A is R6 , number of columns of A.
The co-domain of A is R4 , number of rows of A.
Observe that the words source and target are rather used for the linear mapping T
(see question 3)), the words domain and co-domain are more usual for the matrix.
2) Give an echelon form for A. Give the reduced echelon form of A.
1
1
1
A=
1
2
3
1
2
3
2
0
1
1
1
10
1
1 1 0 0
0
11
0
2
23
0
1
0
0
0
3
1
0
3
3
1
0
0
2
1
0
0
2 110
1
1 1 1 0 0
0 011
0
3 0 0 3
0
1
1
1
0
1
1
0
1
3
1
0
0
2
1
0
0
1
1
0
3
1
1
1
0
0
0
1
0
0
0
0
1
The reduced echelon form is
1
0
0
0
0
1
0
0
1 0 0
1 00
0 10
0 01
2
1
.
0
1
You were expected to show more work than I do.
3) Is the mapping T : x Ax one-to-one?
No it is not because there are non pivot columns, so the homogeneous equation Ax = 0
has non trivial solutions.
Is it onto? (Justify your answers, Yes or No is not acceptable)
Yes it is because there is a pivot in each row.
4) Without further calculation give the Rank of A and the dimension of the null space of
A. (Justify your answers.)
Rank A = 4 is the dimension of Col(A) (or the dimension of the range of T ) that is
the number of pivot columns.
From the rank theorem the dimension of the null space of A is 6 - 4 = 2. It is also
the number of non pivot columns.
5) Determine a basis for the column space of A.
We may take the pivot columns of A, namely:
1
3
1
1
1
1
1
1
A = (a1 = , a2 = , a4 =
, a5 =
)
1
2
0
1
2
3
2
2
(This is what I am expecting. However since T is onto, Col(A) = R4 , and one could
take any basis of R4 , including the standard basis. Be careful! if T is not onto you are to
take the pivot columns of A. No other basis would be natural.)
6) Give the coordinates of each of the columns of A in this basis.
2
1
0
0
0
0
1
0
0
[a1 ]A = , [a2 ]A = , [a4 ]A = , [a5 ]A = .
0
0
1
0
0
0
0
1
To obtain the coordinates of [a3 ]A and [a6 ]A , we are to solve the equations with
augmented matrices: [a1 a2 a4 a5 | a3 ] and [a1 a2 a4 a5 | a6 ].
1
1
1
2
3
1
2
3
1
1
1 1
0
1
2
2
2
1
0 0
1
0
1
0
|
|
|
|
3
1
0
0
1
1
1
0
1
1
1
1
|
|
|
|
2
1
0
0
We may observe that the row reduction has already been achieved in question 2), no
additional calculation was needed.
1
x1 +3x2 +x3 +x4 = 2
x1 = 1
x2
+x3 +x4 = 1
x2 = 1
1
[a3 ]A =
0
x3 +x4 = 0
x3 = 0
x4 = 0
x4 = 0
0
Analogously:
1
1
1
2
x1
+3x2
x2
3
1
2
3
+x3
+x3
x3
1
1
1 1
0
1
2
2
+x4
+x4
+x4
x4
=
=
=
=
0
1
0 0
1
0
3
0
|
|
|
|
x1
x2
x3
x4
0
0
1
1
3
1
0
0
1
1
1
0
=2
= 1
=0
=1
1
1
1
1
|
|
|
|
0
0
1
1
2
1
[a6 ]A =
0
1
II. Show that B = (1 + 2t + 3t2 , 2 + 5t + 7t2 , 3 + 7t + 12t2 ) is a basis for the vector
space of polynomials of degree at most 2. What is the matrix of change of coordinates
PEB from the basis B to the standard basis S ? Give the coordinates of the polynomial
11 + 27t + 42t2 in this basis B.
Using the coordinate vectors of the polynomials in the standard basis of the vector
space of polynomials of degree at most 2, we are to check the linear independence of three
vectors. We write these vectors as columns of a matrix and by row reduction we can
determine that the columns are all pivot columns.
123
123
123
2 5 7 0 1 1 0 1 1
3 7 12
013
002
3
The vectors are linearly and independent so are the polynomials. In the vector space
of polynomials of degree at most 2 which is of dimension 3, three linearly independent
polynomials do provide a basis B.
123
The matrix of change of coordinates is: B = 2 5 7 , we have:
3 7 12
[q ]1,t,t2 = B [q ]B and we are to solve.
The coordinates of the polynomial polynomial 11 + 27t + 42t2 in this basis B are
obtained by solving the linear system with augmented matrix:
123
2 5 7
3 7 12
x1
x
2
x3
1
| 11
| 27 0
| 42
0
2
1
1
3
1
3
|
|
|
123
11
5 0 1 1
001
9
| 11
| 5
|2
=
=
=
11 6 6 = 1
3
. The coordinate vector of the polynomial 11 + 27t + 42t2 in
2
1
this basis B is 3 .
2
2x3
x1
III. 1) Let T x2 = x1 + 2x2 + x3 . Write the standard matrix of T .
x1 + 3x3
x3
0 0 2
The standard matrix of T is 1 2 1 .
0
1 3
0
2
1
0 , 1 , 1 a basis of R3 and why?
2)Is B =
0
1
1
1 0 2
101
10 1
0 1 1 0 1 1 0 1 1
101
1 0 2
0 0 1
There are 3 pivots in a 3x3 matrix: the columns are linearly independent (a pivot in each
column) and span all of R3 (a pivot in each row).
3) What is the matrix of T in the basis B? Show all your work but if you need to get
the inverse of a matrix you can get it directly from your calculator.
x = [x]E = B [x]B , where B is the matrix of
1
0
standard basis. Actually this matrix is B =
1
change coordinates from B to the
of
0 2
1 1 .
01
In B-coordinates T (x) = Ax becomes: B [T (x)]B = AB [x]B .
4
The matrix of T in the basis B is:
[T ]B = B 1 AB.
To get B 1 you may use your calculator or row reduce [B |I ] or use the Adjugate.
IV .
2
We consider the matrix A = 0
3
1) Give the characteristic polynomial of
2
3
0
1
3
4
31
1 1 .
41
A.
1
= (1 )[(2 )(1 ) + 3] [(2 )4 + 9]
1
1
(1 )[2 + 2 + 2 + 3] + 8 + 4 9 = (1 )(2 + + 1) + 4 1 = 3 + 4
2) For a matrix, what is an eigenvalue and what is an associated eigenvector? (Give
denitions)(4points)
is an eigenvalue for a matrix A if there exists a vector x = 0 such that Ax = x.
A vector x = 0 such that Ax = x is an eigenvector associated with the eigenvalue .
3) What are the eigenvalues of A? (4points) Without further calculation can you decide if
A is or is not diagonalizable? (Justify your answer.)(3 points) Is A invertible, and why or
why not? (3points)
The eigenvalues are the roots of the characteristic polynomial 3 + 4. We can
factor: (4 2 ), and the roots are obviously: 0, 2, and -2.
A is diagonalizable because its 3 eigenvalues are distinct.
A is not invertible because 0 is an eigenvalue. So the equation Ax = 0 has non trivial
solutions.
4) What is an eigenspace associated with an eigenvalue? (Give a denition.)(4points)
An eigenspace E associated with the eigenvalue is the set of all x such that Ax = x.
E = {x | Ax = x}
This means that x is either 0 or an eigenvector associated with .
5) Explain without calculation why each of the eigenspaces for A is of dimension 1.(3pts)
For each eigenvalue of A determine a basis for the associated eigenspace and give a basis
B of eigenvectors for R3 .(3points each)
For each of the three eigenvalues, 0 < dimE () = 1. The only possibility is
dimE = 1.
5
I dont have time to give a fully worked out solution. However here are some results.
1
A basis of E0 is B0 = ( 1 )
1
1
A basis of E2 is B2 = ( 1 )
1
5
A basis of E2 is B2 = ( 3 )
9
5
1
1
A basis of eigenvectors is B = 1 , 1 , 3 .
9
1
1
6) Let T : R3 R3 be the linear mapping given in the standard basis by x T (x) = Ax.
By using the matrix of passage P = PEB from the basis B to the standard basis show
that in the basis B the matrix of T is a diagonal matrix D that you will determine. (6
points)
1 1 5
The matrix of change of coordinates from B to the standard basis is P = 1 1 3
119
In the basis B the matrix of T is diagonal,
0
1
0
D = P AP =
0
0
2
0
0
0
2
To avoid the calculation of P 1 we just need to check that AP = P D.But from the
wording of the question you were expected to do this verication.
6