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Here is the given question and my work so far:

Question: Let $A$ be an $n \times n$ invertible matrix, and let $u$ and $v$ be two vectors in $\mathbb{R}^n$. Find the necessary and sufficient conditions on $u$ and $v$ in order that the matrix $$ B=\begin{bmatrix} A & u\\[1ex] v^T & 0 \end{bmatrix} $$ be invertible, and give a formula for the inverse when it exists.

Solution: Performing block Gaussian elimination on $B$, specifically $R_2-v^T A^{-1}R_1\longrightarrow R_2$, we obtain

$$ B=\begin{bmatrix} A & u\\[1ex] 0 & -v^TA^{-1}u \end{bmatrix} $$

Next, recall that in order for a matrix to be nonsingular, it is both necessary and sufficient that its determinant be nonzero. Before we proceed, take note of the fact that from $(-v^T A^{-1} u)$ we produce $(1\times n) \cdot (n\times n) \cdot (n\times 1) = 1\times 1$. Thus, the product produces a scalar.

Now, we wish to show that $\det(B)\neq 0$. Then,

$$\det (B) = \det (A\cdot -v^T A^{-1} u)=(-v^T A^{-1} u)\cdot \det(A) $$

Because $A$ is nonsingular, we know that $\det(A)\neq 0$ and moreover, that $A^{-1}u\neq0$ for $u\neq 0$.

My Question: How do I show that $v^T A^{-1} u\neq 0$?

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  • $\begingroup$ The matrix is invertible if and only if $v^TA^{-1}u\ne0$. Do Laplace expansion on the last row. $\endgroup$ – egreg Apr 1 '16 at 21:36
  • $\begingroup$ @egreg Wouldn't that just produce what I already have? Namely, $\det (B) = \det (A\cdot -v^T A^{-1} u)=(-v^T A^{-1} u)\cdot \det(A) $ $\endgroup$ – hungryformath Apr 1 '16 at 21:41
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You did everything right (up to the computation of the determinant, which is not written down correctly). Doing Laplace expansion on the last row, $$ \det B=(-v^TA^{-1}u)\det A $$ so this is nonzero if and only if $v^TA^{-1}u\ne0$.

You can't prove that $v^TA^{-1}u\ne0$, because it's false in general (think to $u=0$, for instance), but the question was asking precisely to show an “if and only if” statement.

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  • $\begingroup$ Ok. From the given question "Find the necessary and sufficient conditions on $u$ and $v$ such that $B$ is invertible". Would stating that $v^T A^{-1} u\neq 0$ be the nasc? The only reason I'm feeling apprehensive about that is because it involves $A^{-1}$. $\endgroup$ – hungryformath Apr 1 '16 at 21:57
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    $\begingroup$ @hungryformath You can't really say more than that: in terms of the inner product, this means $v$ is not orthogonal to $A^{-1}u$, so the condition does depend on $A$. $\endgroup$ – egreg Apr 1 '16 at 22:00

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