## 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 $L_1 = L_2 \quad D_1 = D_2 \quad U_1 = U_2$ $L_1D_1U_1 = L_2D_2U_2 \rightarrow L_1^{-1}L_1D_1U_1 = L_1^{-1}L_2D_2U_2 \rightarrow D_1U_1 = L_1^{-1}L_2D_2U_2$ $\rightarrow D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2U_2U_2^{-1} \rightarrow D_1U_1 U_2^{-1} = L_1^{-1}L_2D_2$

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 .

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