A matrix version of L'Hopital's Rule?

Question

Is there a version of L'Hopital's Rule for matrix calculus?

For example: let $A$ be a symmetric $n\times n$ positive definite matrix and $b$ be an $n\times 1$ vector. As $b$ converges to $0_{n\times 1}$,

$$b'Ab \to 0, $$ and $$Abb'A \to 0_{n\times n}.$$

Find $$\lim_{b\to 0} \frac{Abb'A}{b'Ab}$$or prove that it does not exist.

Thanks.

Note: $b'$ denotes the transpose of $b$: $b' \equiv b^{T} $.

OP's using prime to denote transpose, a la Matlab, I believe. — John Hughes, Dec 22 '14 at 21:00
@JohnHughes, just FYI, the prime operator is actually the Hermetian in MATLAB. True transpose (that works on complex numbers) is denoted ".'". — FundThmCalculus, Dec 22 '14 at 21:05
Since we're talking about positive definite, hence real, matrices, transpose and transpose-conjugate are the same. :) But thanks for the matlab tip. — John Hughes, Dec 22 '14 at 21:15

score 2 · Accepted Answer · answered Dec 22 '14 at 21:13

The limit does not exist. Consider two orthonormal eigenvectors $v_1, v_2$ of $A$, and first let's look at

$$\lim_{t\to 0} \frac{A(tv_1)(tv_1)^TA}{tv_1^TAtv_1} = \lim_{t\to 0} \frac{t^2 \lambda_1^2 v_1 v_1^T } {t^2\lambda_1} = \lambda_1 v_1v_1^T$$

and compare to

$$\lim_{t\to 0} \frac{A(tv_2)(tv_2)^TA}{tv_2^TAtv_2} = \lambda_2 v_2v_2^T.$$

These cannot be equal, as can be checked by multiplying both by $v_1$:

$$\lambda_1 v_1 v_1^Tv_1 = \lambda_1 v_1 \neq 0 = \lambda_2 v_2 v_2^Tv_1.$$

1 Answers1