Section ROD  Rank One Decomposition

From A First Course in Linear Algebra
Version 2.20
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

This Section is a Draft, Subject to Changes

Our first decomposition applies only to diagonalizable (Definition DZM) matrices, and yields a decomposition into a sum of very simple matrices.

Theorem ROD
Rank One Decomposition
Suppose that A is a diagonalizable matrix of size n and rank r. Then there are r square matrices A1,A2,A3,,Ar, each of size n and rank 1 such that

A = A1 + A2 + A3 + + Ar

Furthermore, if λ1,λ2,λ3,,λr are the nonzero eigenvalues of A, then there are two sets of r linearly independent vectors from n,

X = x1,x2,x3,,xr Y = y1,y2,y3,,yr

such that Ak = λkxkykt, 1 k r.

Proof   The proof is constructive. Generally, we will diagonalize A, creating a nonsingular matrix S and a diagonal matrix D. Then we split up the diagonal matrix into a sum of matrices with a single nonzero entry (on the diagonal). This fundamentally creates the decomposition in the statement of the theorem, the remainder is just bookkeeping. The vectors in X and Y will result from the columns of S and the rows of S1.

Let λ1,λ2,λ3,,λn be the eigenvalues of A (repeated according to their algebraic multiplicity). If A has rank r, then dim NA = n r (Theorem RPNC). The null space of A is the eigenspace of the eigenvalue λ = 0 (Theorem EMNS), so it follows that the algebraic multiplicity of λ = 0 is n r, αA 0 = n r. Presume that the complete list of eigenvalues is ordered so that λk = 0 for r + 1 k n.

Since A is hypothesized to be diagonalizable, there exists a diagonal matrix D and an invertible matrix S, such that D = S1AS. We can rearrange tis equation to read, A = SDS1. Also, the proof of Theorem DC says that the diagonal elements of D are the eigenvalues of A and we have the flexibility to assume they lie on the diagonal in the same order as we have specified above. Now, let X = x 1,x2,x3,,xn be the columns of S, and let Y = y 1,y2,y3,,yn be the rows of S1 converted to column vectors. With little motivation other than the statement of the theorem, define size n matrices Ak, 1 k n by Ak = λkxkykt. Finally, let Dk be the size n matrix that is totally zero, other than having λk in row k and column k.

With everything in place, we compute entry-by-entry,

Aij = SDS1 ij  Definition DZM = S k=1nD k S1 ij  Definition MA = S k=1nD kS1 ij  Theorem MMDAA = k=1nSD kS1 ij  Theorem MMDAA = k=1n SD kS1 ij  Definition MA = k=1n =1n SD k i S1 j  Theorem EMP = k=1n =1n p=1n S ip Dk p S1 j  Theorem EMP = k=1n S ik Dk kk S1 kj   Dk p = 0 if pk, or k = k=1n S ikλk S1 kj Dk kk = λk = k=1nλ k Sik S1 kj  Property CMCN = k=1nλ k xk i1 ykt 1j  Definition of XY = k=1nλ k q=11 x k iq ykt qj = k=1nλ k xkykt ij  Theorem EMP = k=1n λ kxkykt ij  Definition MSM = k=1n A k ij  Definition of Ak = k=1nA k ij  Definition MA

So by Definition ME we have the desired equality of matrices. The careful reader will have noted that Ak = O, r + 1 k n, since λk = 0 in these instances. To get the sets X and Y from X and Y , simply discard the last n r vectors. We can safely ignore (or remove) Ar+1,Ar+2,,An from the summation just derived.

One last assertion to check. What is the rank of Ak, 1 k r? Every row of Ak is a scalar multiple of ykt, row k of the nonsingular matrix S1 (Theorem MIMI). As a row of a nonsingular matrix, ykt cannot be all zeros. In particular, row i of Ak is obtained as a scalar multiple of ykt by the scalar αk xk i. We have restricted ourselves to the nonzero eigenvalues of A, and as S is nonsingular, some entry of xk is nonzero. This all implies that some row of Ak will be nonzero. Now consider row-reducing Ak. Swap the nonzero row up into row 1. Use scalar multiples of this row to zero out every other row. This leaves a single nonzero row in the reduced row-echelon form, so Ak has rank one.

We record two observations that was not stated in our theorem above. First, the vectors in X, chosen as columns of S, are eigenvectors of A. Second, the product of two vectors from X and Y in the opposite order, by which we mean yitx j, is the entry in row i and column j of the matrix product S1S = I n (Theorem EMP). In particular,

yitx j = 1 if i = j 0  if ij

We give two computational examples. One small, one a bit bigger.

Example ROD2
Rank one decomposition, size 2
Consider the 2 × 2 matrix,

A = 166 45 17

By the techniques of Chapter E we find the eigenvalues and eigenspaces,

λ1 = 2 A 2 = 1 3 λ2 = 1 A 1 = 2 5

With n = 2 distinct eigenvalues, Theorem DED tells us that A is diagonalizable, and with no zero eigenvalues we see that A has full rank. Theorem DC says we can construct the nonsingular matrix S with eigenvectors of A as columns, so we have

S = 12 3 5 S1 = 5 2 31

From these matrices we obtain the sets of vectors

X = 1 3 , 2 5 Y = 5 2 , 3 1

And we have the matrices,

A1 = 2 1 3 5 2 t = 2 52 15 6 = 104 30 12 A2 = (1) 2 5 3 1 t = (1) 6 2 155 = 62 15 5

And you can easily verify that A = A1 + A2.

Here’s a slightly larger example, and the matrix does not have full rank.

Example ROD4
Rank one decomposition, size 4
Consider the 4 × 4 matrix,

B = 34 18 1644 24 1 9 36 18 36 36 18 6 3

By the techniques of Chapter E we find the eigenvalues and eigenvectors,

λ1 = 3 B 3 = 1 2 1 1 , 1 1 1 2 λ2 = 2 B 2 = 1 2 0 0 λ3 = 0 A 0 = 2 3 2 2

The algebraic and geometric multiplicities of each eigenvalue are equal, so Theorem DMFE tells us that A is diagonalizable. With a single zero eigenvalue we see that A has rank 4 1 = 3. Theorem DC says we can construct the nonsingular matrix S with eigenvectors of A as columns, so we have

S = 1 1 1 2 2 1 2 3 1 1 0 2 1 2 0 2 S1 = 4 2 0 1 8 4 1 1 1 0 1 0 6 3 1 1

Since r = 3, we need only collect three vectors from each of these matrices,

X = 1 2 1 1 , 1 1 1 2 , 1 2 0 0 Y = 4 2 0 1 , 8 4 11 , 1 0 1 0

And we obtain the matrices,

B1 = 3 1 2 1 1 4 2 0 1 t = 3 4 2 018 4 0 2 4 2 014 2 0 1 = 12 6 0324 12 0 6 12 6 0312 6 0 3 B2 = 3 1 1 1 2 8 4 11 t = 3 8 4 118 4 1 1 8 4 11 16 8 2 2 = 24 12 3324 12 3 3 24 12 33 48 24 6 6 B3 = (2) 1 2 0 0 1 0 1 0 t = (2) 1 0102 0 2 0 0 0 0 0 0 0 0 0 = 20 2 0 4 0 4 0 0 0 0 0 0 0 0 0

Then we verify that

B = B1 + B2 + B3 = 12 6 0324 12 0 6 12 6 0312 6 0 3 + 24 12 3324 12 3 3 24 12 33 48 24 6 6 + 20 2 0 4 0 4 0 0 0 0 0 0 0 0 0 = 34 18 1644 24 1 9 36 18 36 36 18 6 3