Prove $A^2 = A$ where $A = I_n − \alpha\alpha^T$ and $\alpha$ is an $n × 1$ vector with $\alpha^T\alpha=1$

Consider a $$n × n$$ matrix $$A = I_n − \alpha\alpha^T$$, where $$I_n$$ is the $$n × n$$ identity matrix and $$α$$ is an $$n × 1$$ column vector such that $$\alpha^T\alpha = 1$$. Show that $$A^2 = A$$.

My proof:

$$\alpha\alpha^T$$ is a $$n × n$$ matrix. It is given that $$\alpha^T\alpha = 1$$. Multiplying both sides of this by the inverse of $$\alpha^T$$ gives, $${{\alpha^T}^{-1}}\alpha^T={\alpha^T}^{-1}I$$.

$$I\alpha={\alpha^T}^{-1}$$ which means $$\alpha={\alpha^T}^{-1}$$.

Now, $$A = I_n − \alpha\alpha^T$$, =$$I_n - {\alpha^T}^{-1}{\alpha^T}$$ which means $$A=I_n-I_n=\mathsf{O}$$.

Therefore, $$\forall n \in \mathbb{N}, A^n =\mathsf{O}$$.

Is it correct?

• Start from $A^2=(I_n-\alpha\alpha^T)(I_n-\alpha\alpha^T)=\cdots$. – StubbornAtom May 2 at 10:11
• You should use \alpha for typesetting $\alpha$, and similarly for other Greek letters. – StubbornAtom May 2 at 10:16

$$\alpha^{T}$$ is a vector. Inverse of a vector does not make sense. Just calculate $$A^{2}$$: $$A^{2}=I-2\alpha \alpha^{T}I +\alpha \alpha^{T}\alpha \alpha^{T}I$$. Since $$\alpha (\alpha^{T}\alpha) \alpha^{T}=\alpha \alpha^{T}$$ by hypothesis we get $$A^{2}=A$$.

$$(I_n-\alpha\alpha^T)^2 = (I_n-\alpha\alpha^T)(I_n-\alpha\alpha^T) = I_n-2\alpha\alpha^T+\alpha\alpha^T\alpha\alpha^T = I_n-2\alpha\alpha^T+\alpha1\alpha^T = I_n-\alpha\alpha^T$$

• @MariaMazur the accepted answer might have been preferred by the OP because there is also some attention for the effort of the OP. – drhab May 2 at 11:01

Observe that: $$A\alpha=\alpha-\alpha\alpha^T\alpha=0\tag1$$ since $$\alpha^T\alpha=1$$.

Further for an arbitrary $$n\times1$$ vector $$v$$: $$Av=v-\alpha\alpha^Tv=v-c\alpha\tag2$$where $$c:=\alpha^Tv$$.

Combining $$(1)$$ and $$(2)$$ we find that:$$A^2v=A(v-c\alpha)=Av$$

This proves that $$A^2=A$$.