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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?

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

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

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    $\begingroup$ @MariaMazur the accepted answer might have been preferred by the OP because there is also some attention for the effort of the OP. $\endgroup$ – drhab May 2 at 11:01
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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 $A^2=A$.

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