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Exam3reviewThe exam will cover the material through the singular value decomposition.Linear transformations and change of basis will be covered on the final.

The main topics on this exam are:

Eigenvalues and eigenvectorsDifferential equations d

dtu =Auand exponentials eAtSymmetric matrices A =AT: These always have real eigenvalues, andthey always have enough eigenvectors. The eigenvector matrix Qcanbe an orthogonal matrix, with A=QQT.Positive definite matricesSimilar matrices B =M1AM . Matrices Aand Bhave the same eigen- values; powers ofAwill look like powers ofB.

Singular value decompositionSampleproblems

1. This is a question about a differential equation with a skew symmetricmatrix.

Suppose 0 1 0

0du

u.Au= 1 1

0=

dt 0 1The general solution to this equation will look like

u(t) =c1e1tx1 +c2e2tx2 +cee3tx3.a) What are the eigenvalues ofA?

The matrix A is singular; the first and third rows are dependent, soone eigenvalue is 1 =0. We might also notice that Ais antisymmet-ric (AT =A) and realize that its eigenvalues will be imaginary.To find the other two eigenvalues, well solve the equation |AI =|0.

1 01 =31

2=0.

0 1

We conclude 2 =2 iand 3 =2 i.

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At this point we know that our solution will look like:

u(t) =c1x1 +c2e2 itx2 +cee2 itx3.We can now see that the solution doesnt increase without bound or de-cay to zero. The size ofei is the same for any ; the exponentials herecorrespond to points on the unit circle.

b) The solution is periodic. When does it return to its original value?(What is its period?)

This is not likely to be on the exam, but we can quickly remark that

e

2 it =e0 when 2t=2, or when t=2.c) Show that two eigenvectors ofAare orthogonal.

The eigenvectors of a symmetric matrix or a skew symmetric matrix

are always orthogonal. One choice of eigenvectors ofAis:

x1 = 01 , x2 = 2

1

i , x3 = 21

i .1 1 1

Dont forget to conjugate the first vector when computing the innerproduct of vectors with complex number entries.

d) The solution to this differential equation is u(t) = eAtu(0). Howwould we compute eAt?

IfA=SS1 then eAt =SetS1 where e1t

t . e=

..

.

ent

So eAt comes from the eigenvalues in and the eigenvectors in S.Fact: A matrix has orthogonal eigenvectors exactly when AAT = ATA;

i.e. when A commutes with its transpose. This is true of symmetric, skewsymmetric and orthogonal matrices.

2. Were told that a three by three matrix Ahas eigenvalues 1 =0, 2 =cand 3 =2 and eigenvectors

1 1 1x1 = 1 , x2 =

1 , x3 = 1 .1 0 2

a) For which cis the matrix diagonalizable?The matrix is diagonalizable if it has 3 independent eigenvectors. Notonly are x1, x2 and x3 independent, theyre orthogonal. So the matrixis diagonalizable for all values ofc.

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b) For which values ofcis the matrix symmetric?IfA=QQT is symmetric its eigenvalues (the entries of) are real.On the other hand, ifcis real, then AT =QTTQ=Ais symmetric.The matrix is symmetric for all real numbers c.

c) For which values ofcis the matrix positive definite?All positive definite matrices are symmetric, so cmust be real. Theeigenvalues of a positive definite matrix must be positive. The eigen-value 0 is not positive, so this matrix is not positive definite for anyvalues ofc. (Ifc0 then the matrix is positive semidefinite.)

d) Is it a Markov matrix?

In a Markov matrix, one eigenvalue is 1 and the other eigenvalues aresmaller than 1. Because 3 = 2, this cannot be a Markov matrix forany value ofc.

e) Could P= 21Abe a projection matrix?Projection matrices are real and symmetric so their eigenvalues arereal. In addition, we know that their eigenvalues are 1 and 0 because

P2 =Pimplies 2 =. So 21Acould be a projection matrix ifc=0 orc=2.

Note that it was the fact that the eigenvectors were orthogonal that madeit possible to answer many of these questions.

Singular value decomposition (SVD) is a factorization

A=(orthogonal)(diagonal)(orthogonal) =UVT.We can do this for any matrix A. The key is to look at the symmetric matrixATA = VTVT; here V is the eigenvector matrix for ATAand T is thematrix of eigenvalues i2 ofATA. Similarly, AAT = UTUT and U is theeigenvector matrix for AAT. (Note that we can introduce a sign error if wereunlucky in choosing eigenvectors for the columns ofU. To avoid this, use theformula Avi =iui to calculate Ufrom V.)

On the exam, you might be asked to find the SVD of a matrix Aor youmight be given information on U, and Vand asked about A.

3 03. Suppose =

0 2and Uand Veach have two columns.

a) What can we say about A?

We know A is a two by two matrix, and because U, Vand are allinvertible we know Ais nonsingular.

3 0b) What if=

0 5 ?This is not a valid possibility for . The singular values the diagonalentries of are never negative in a singular value decomposition.

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3 0c) What if=

0 0?

Then Ais a singular matrix of rank 1 and its nullspace has dimension1. The four fundamental subspaces associated with Aare spanned byorthonormal bases made up of selected columns ofUand V. In thisexample, the second column ofVis a basis for the nullspace ofA.

4. Were told that Ais symmetric and orthogonal.a) What can we say about its eigenvalues?

The eigenvalues of symmetric matrices are real. The eigenvalues oforthogonal matrices Qhave ||=1; multiplication by an orthogonalmatrix doesnt change the length of a vector. So the eigenvalues ofAcan only be 1 or 1.

b) True or false: Ais sure to be positive definite.

1 0False it could have an eigenvalue of1, as in 1 .0

c) True or false: Ahas no repeated eigenvalues.False ifAis a three by three matrix or larger, its guaranteed to haverepeated eigenvalues because every is 1 or 1.

d) Is Adiagonalizable?Yes, because all symmetric and all orthogonal matrices can be diago-nalized. In fact, we can choose the eigenvectors ofAto be orthogonal.

e) Is Anonsingular?Yes; orthogonal matrices are all nonsingular.

f) Show P= 21 (A+I)is a projection matrix.We could check that Pis symmetric and that P2 =P:

2P2 = 1 (A+I) = 1 (A2 +2A+I).

2 4

Because Ais orthogonal and symmetric, A2 =ATA=I, soP2 = 1 (A2 +2A+I) = 1 (A+I) =P.

4 2

Or we could note that since the eigenvalues ofAare 1 and 1 thenthe eigenvalues of12 (A+I)must be 1 and 0.

These questions all dealt with eigenvalues and special matrices; thats whatthe exam is about.

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18.06SC Linear AlgebraFall 2011

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