## Linear Algebra and Its Applications, Review Exercise 1.25

Review exercise 1.25. Given the matrix $A$ where

$A = \begin{bmatrix} 1&0&0 \\ 2&1&0 \\ 0&5&1 \end{bmatrix} \begin{bmatrix} 1&2&0 \\ 0&1&5 \\ 0&0&1 \end{bmatrix}$

what multiple of row 2 was subtracted from row 3 in elimination? Explain why $A$ is invertible, symmetric, and tridiagonal. What are the pivots?

Answer: From the above we see that $A$ has been factored into $A = LL^T$. Since $L$ contains the multipliers used in elimination, we can look at $l_32 = 5$ to determine that 5 times row 2 was subtracted from row 3 in elimination.

Since $A$ can be factored into the product of a lower triangular matrix $L$ and upper triangular matrix $L^T$ we know that $A$ is invertible. Since $A = LL^T$ we have $A^T = (LL^T)^T = (L^T)^TL^T = LL^T = A$ so $A$ is also symmetrical. Since $L$ and $L^T$ are bidiagonal the original matrix $A$ was tridiagonal. Finally, since we can express $A$ as $A = LDL^T$ where $D = I$ we see that the pivots (i.e., the diagonal entries of D) are all 1.

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.