Linear Algebra and Its Applications, Exercise 1.6.18

Exercise 1.6.18. Suppose that

A = L_1D_1U_1 = L_2D_2U_2

Show that

L_1 = L_2 \quad D_1 = D_2 \quad U_1 = U_2

Answer: We have

L_1D_1U_1 = L_2D_2U_2 \rightarrow L_1^{-1}L_1D_1U_1 = L_1^{-1}L_2D_2U_2 \rightarrow D_1U_1 = L_1^{-1}L_2D_2U_2

\rightarrow D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2U_2U_2^{-1} \rightarrow D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2

The product of two upper triangular matrices is also an upper triangular matrix, and multiplying by a diagonal matrix preserves this. The left side of the final equation above is therefore an upper triangular matrix. Similarly the product of two lower triangular matrices is also an lower triangular matrix, and multiplying by a diagonal matrix does not change this. The right side of the final equation above is therefore an lower triangular matrix.

In order for both of these things to be true, the left and right hand sides of the final equation above must therefore both equal some diagonal matrix D:

D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2 = D

Now, since D_1 is a diagonal matrix and D is also a diagonal matrix, the product U_1U_2^{-1} must also be a diagonal matrix. If it were not, and had at least one entry

u_{ij} \ne 0, i \ne j

then the (i, j) entry of D would be equal to the product of the nonzero (i, i) entry of D_1 and the nonzero (i, j) entry of U_1U_2^{-1} and would therefore be nonzero itself. But this is a contradiction since D is a diagonal matrix, so the product U_1U_2^{-1} must be a diagonal matrix.

Also, by definition U_1 and U_2 have ones on the diagonal. Since U_2 is an upper triangular matrix with ones on the diagonal, its inverse (which is also an upper triangular matrix, as shown in exercise 1.16.12) must have ones on the diagonal as well, as shown by the following argument:

Let U_2 have entries u_{ij} and U_2^{-1} have entries v_{ij}. Since U_2U_2^{-1} = I, for the (i, i) entry of U_2U_2^{-1} we therefore have

\sum_k u_{ik}v_{ki} = 1

But since both U_1 and U_2^{-1} are upper triangular we have

u_{ik} = 0, k < i

and

v_{ki} = 0, k > i

so that

1 = \sum_k u_{ik}v_{ki} = u_{ii}v_{ii} = v_{ii}

since U_1 has ones on the diagonal.

Since both U_1 and U_2^{-2} are upper triangular matrices with ones on the diagonal, their product U_1U_2^{-1} also has ones on the diagonal. But we also know that this product is a diagonal matrix, so that we then have

U_1U_2^{-1} = I \rightarrow U_1 = U_2

since U_1 has a unique inverse.

A similar argument shows that

L_1^{-1}L_2 = I \rightarrow L_1 = L_2

We then have

D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2 \rightarrow D_1I = ID_2 \rightarrow D_1 = D_2

and the factorization A = LDU is therefore unique.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s