show inverse by matrix multiplication

Question

Suppose $v^Tu \neq 1$ and $u,v \in \mathbb{R}^n$. Both $u$ and $v$ are column vectors. Define matrix $A=I+uv^T$. Show by matrix multiplication that $$A^{-1}=I-\frac{uv^T}{1-v^Tu}$$

My attempt: $$AA^{-1}=(I+uv^T)(I-\frac{uv^T}{1-v^Tu})=I-\frac{uv^T}{1-v^Tu} +uv^T-\frac{uv^Tuv^T}{1-v^Tu}=$$ $$I-\frac{1}{1-v^Tu}(uv^T-uv^T(1-v^Tu)-uv^Tuv^T)$$

I got stuck at part shown above. Can anyone help me?

you should have $\dfrac{uv^Tuv^T}{1-v^Tu}$ instead of $\dfrac{uv^T}{1-v^Tu}$ — Petite Etincelle, Sep 22 '14 at 10:26

score 2 · Accepted Answer · answered Sep 22 '14 at 10:34

2

\begin{align} (I + uv^T)(I-\dfrac{uv^T}{1+v^Tu}) &= I -\dfrac{uv^T}{1+v^Tu} +uv^T-\dfrac{uv^Tuv^T}{1+v^Tu} \\&= I-\dfrac{uv^T}{1+v^Tu} +uv^T-v^Tu\dfrac{uv^T}{1+v^Tu} \\ &=I + uv^T -(1+v^Tu)\dfrac{uv^T}{1+v^Tu}\\ & = I \end{align}

So the right answer is $A^{-1} = I-\dfrac{uv^T}{1+v^Tu}$

answered Sep 22 '14 at 10:34

Petite Etincelle

14,767

Can you explain second and third equalities? – Idonknow Sep 22 '14 at 10:41
@Idonknow $uv^Tuv^T = u(v^Tu)v^T$ and as you said in the comment $v^Tu$ is a real number, so $u(v^Tu)v^T = (v^Tu)uv^T$ – Petite Etincelle Sep 22 '14 at 10:43

show inverse by matrix multiplication

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