Introduction Matrix Action Perpframes, Aligners and Hangers Stretchers Coordinates Projections SVD Matrix Subspaces Linear Systems, PseudoInverse Condition Number Matrix Norm, Rank One Data Compression Noise Filtering
Todd Will

Frobenius
Norm
Rank One Decomposition 
Matrix Norm andRank One Decomposition 
Lower
Rank Approximation
Exercises 
Frobenius Norm.The singular values of the matrix are .When you add up the squares of each entry of A you get
When you add the squares of the singular values you get
Accident? In math there are no accidents! The Frobenius norm of a matrix A, , is defined as the square root of the sum of the squares of all its entries. E.g.
Theorem: If A has singular values , then .Proof:Let be an SVD of A. First note that for any matrix
given in terms of its columns,
Now,
(1) Let be the columns of (stretcher)(aligner). Since the hanger matrix simply rotates the columns of without changing their lengths, the two sides are equal. (2) The matrix simply rotates the columns of without changing their lengths.
Rank One DecompositionSuppose the SVD for a matrix isYou've seen that the reduced SVD is Here's one more way to write A, called the rank one decomposition.
How do you know this equals ?
As usually it's the agreement on the basis .
Here's what happens when the rank one decomposition hits :
(Linearity) (Since is orthonormal)
Since the rank one decomposition agrees with A on , it must equal A. The rank one decomposition gets its name from that fact that each of the summand is a rank one matrix.
Lower Rank ApproximationsHere's a 3x3 matrix .When you compute the determinant of A you find Det[A]=16, so you know that A has rank three. This big fat juicy determinants tells you that for every y, the system A x = y, has the unique solution . Now check out the matrix B and note that A is approximately equal to B. In fact you can compute
But when you check the determinant of B you find Det[B]=0. A zero determinant tells you that for some y's, B x = y, will
have infinitely many solutions and for other y's there'll be no
solutions at all.
Is there a way to predict that the seemingly wellbehaved matrix A has
such a nasty neighbor B so nearby?
You bet. Here's an SVD for A The three nonzero singular values tell you that the matrix has rank 3. But the value 0.01 is so small that A is nearly a rank two matrix. In fact the matrix B was created by setting that last singular value
to zero.
Now the rank one decomposition of A is
and the rank one decomposition of B is
So
and .
So you see that if A has a small singular value, then you can get a lower rank matrix B close to A by setting the small singular value to zero. The next theorem says that this is the best way to find nearby neighbors of lower rank.
Theorem: Suppose A is an mxn matrix with singular values . If A has rank r, then for any k<r, a best rank k approximation to A is .For a proof of this theorem see page 414 of Steven J. Leon's excellent book, Linear Algebra with Applications, 5th Edition.Comments: Best means that is minimized. Note that
.
Exercises1. Compute2. The first two singular values of
are
and .
