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We consider the matrix $A=I-a\vec{v}\vec{v}^T$, where $I$ is the $n\times n$ identity matrix and $\vec{v}$ is a unit vector $n\times 1$. Find for which value of $a$ the determinant of the matrix $A$ is zero. If the determinant of the matrix $A$ is not zero, find the value of $b$ such that $I+b\vec{v}\vec{v}^T$ is the inverse of $A$. For $a>0$ apply this for the inversion of the matrix $A=\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix}$.

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Could you give me a hint for the first part, about how to compute $a$ ? To what is $\det (I-a\vec{v}\vec{v}^T)$ equal?

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As for the second part:

We want that $A^{-1}=I+b\vec{v}\vec{v}^T$.

We have the following:\begin{align*}A\cdot A^{-1}=I &\Rightarrow (I-a\vec{v}\vec{v}^T)\cdot (I+b\vec{v}\vec{v}^T)=I \\ & \Rightarrow I(I+b\vec{v}\vec{v}^T)-a\vec{v}\vec{v}^T(I+b\vec{v}\vec{v}^T)=I \\ & \Rightarrow I+b\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=I \\ & \Rightarrow b\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=0 \\ & \Rightarrow b\vec{v}\vec{v}^T-ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T=a\vec{v}\vec{v}^T \\ & \Rightarrow b\left (\vec{v}\vec{v}^T-a\vec{v}\vec{v}^T\vec{v}\vec{v}^T\right )=a\vec{v}\vec{v}^T \\ & \Rightarrow b\left (I-a\vec{v}\vec{v}^T\right )\vec{v}\vec{v}^T=a\vec{v}\vec{v}^T\\ & \Rightarrow b\left (I-a\vec{v}\vec{v}^T\right )=a\\ & \Rightarrow b=a\left (I-a\vec{v}\vec{v}^T\right )^{-1}\end{align*}

Is this correct?

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As for the last part:

We have that \begin{align*}A&=\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix} =\begin{pmatrix}1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \ldots & 1\end{pmatrix}+\begin{pmatrix}a & a & \ldots & a \\ a & a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & a\end{pmatrix} =I+a\begin{pmatrix}1 & 1 & \ldots & 1 \\ 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \vdots & \vdots \\ 1 & 1 & \ldots & 1\end{pmatrix} \\ & =I+a\begin{pmatrix}1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}\begin{pmatrix}1 & 1 & \ldots & 1\end{pmatrix} =I-a\begin{pmatrix}-1 \\ -1 \\ \vdots \\ -1 \end{pmatrix}\begin{pmatrix}-1 & -1 & \ldots & -1\end{pmatrix}\end{align*}

So we have in this case $\vec{v}=\begin{pmatrix}-1 \\ -1 \\ \vdots \\ -1 \end{pmatrix}$.

Therefore the inverse matrix is $A^{-1}=I+b\vec{v}\vec{v}^T$ with $b=a\left (I-a\vec{v}\vec{v}^T\right )^{-1}$.

Is this part correct?

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4 Answers 4

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The linear operator $A = I - a v v^T$ is quite simple. You can check that $Av = (1-a)v$ and so $v$ is an eigenvector, whilst for any vector $w$ perpendicular to $v$ we have $Aw = w$. Therefore there is a one-dimensional eigenspace with eigenvalue $(1-a)$ spanned by $v$, and an $(n-1)$-dimensional eigenspace with eigenvalue $1$ consisting of all vectors perpendicular to $v$.

This should answer all your questions, for instance the determinant should be the product (with multiplicity) of eigenvalues.

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Add columns $\;C_2, C_3,...,C_n\;$ to the first one $\;C_1\;$ ,and get:

$$\begin{pmatrix}1+na&a&\ldots&a\\ 1+na&1+a&\ldots&a\\ \ldots&\ldots&\ldots&\ldots\\ 1+na&a&\ldots&1+a\end{pmatrix}$$

Now, the Gauss operatios $\;C_k-C_1\;,\;\;k=2,3,...,n\;$ :

$$\begin{pmatrix}1+na&a&\ldots&a\\ 0&1&\ldots&0\\ \ldots&\ldots&\ldots&\ldots\\ 0&0&\ldots&1\end{pmatrix}$$

Observe that the operations do not affect the value of the determinant. Now, calculate the determinant value in the last matrix by the first column...

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  • $\begingroup$ Is the form of the matrix $A=I-a\vec{v}\vec{v}^T$ always $\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix}$ no matter what $\vec{v}$ is? $\endgroup$
    – Mary Star
    Mar 27, 2020 at 11:23
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$\frac{1}{a}A=\frac{1}{a}I-vv^T$ where $a\ne 0$

$\implies det(\frac{1}{a}A)=det(\frac{1}{a}I-vv^T)$

You wish $det(\frac{1}{a}I-vv^T)=0$ which implies the matrix $vv^T$ must have an eigenvalue equal to $1/a$.

For an $n\times n$ matrix $vv^T$ the eigenvalues are:

$trace (vv^T)(=1)$ and $0$ (with algebraic multiplicity $n-1$). This suggests that $1/a=1\implies a=1$.

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  • $\begingroup$ Why are the only possible eigenvalues $1$ and $0$ ? $\endgroup$
    – Mary Star
    Mar 27, 2020 at 13:59
  • $\begingroup$ Let $v=(x_1,x_2,...,x_n)^T$ then characteristic equation of $vv^T$ is $t^n-||v||^2t^{n-1}=0$ where $||v||=\sqrt{x_1^2+x_2^2+....+x_n^2}$. $\endgroup$ Mar 27, 2020 at 14:07
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The second part is wrong, because $b$ is a number while $a(I-a\vec{v}\vec{v}^T)^{-1}$ is a matrix.

Note that $\vec{v}$ is a unit vector, this means $\vec{v}^T\vec{v} = 1.$ You can use this in $ab\vec{v}\vec{v}^T\vec{v}\vec{v}^T$ and get $(b-a-ab)\vec{v}\vec{v}^T = 0$ or $b=\frac{a}{1-a}.$

This allows you to answer the first part: If $a\neq 1,$ you can find the inverse of $A,$ which means that $\det(A)\neq 0.$ If $a=1,$ then $A\vec{v} = (I-\vec{v}\vec{v}^T)\vec{v} = \vec{v}-\vec{v}= \vec{0},$ which means that there is a non-zero vector that maps to $\vec{0},$ which in turn means $\det(A)=0.$

Regarding the exact value of the determinant: If you are familiar with eigenvalues and eigenvectors, I recommend to take a look at Joppy's answer. Without knowledge about eigenvalues and eigenvectors, this is going to be tricky. I cannot come up with a simple solution (not using at least the matrix determinant lemma) at the moment. But as far as I can see, this has not been asked in the original question.

The last part is incorrect, too. When you change $\vec{v}$ from $(1\;\ldots\; 1)^T$ to $(-1\;\ldots\; -1)^T,$ the matrix $A$ does not change, because the sign cancels out in $\vec{v}\vec{v}^T.$ Furthermore, $\vec{v}$ is not a unit vector anymore.

You should rename the "$a$" in the formula $(I-a\vec{v}\vec{v}^T)^{-1} = I+\frac{a}{1-a}\vec{v}\vec{v}^T$ in order not to confuse it with the "$a$" in the last part of the exercise. Let's say $(I-c\vec{v}\vec{v}^T)^{-1} = I+\frac{c}{1-c}\vec{v}\vec{v}^T$

Now you have to figure out which value $c$ and which vector $\vec{v}$ to choose in order to obtain the given matrix, while $\vec{v}$ is subject to the constraint $\vec{v}^T\vec{v}=1.$ It turns out that $\vec{v} = \sqrt{\frac{1}{n}} (1 \; \ldots \; 1)^T$ and $c= -na.$

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  • $\begingroup$ Thank you very much for your answer!! As for the last part: We get the following inverse matrix: \begin{align*}A^{-1}&=I+\frac{c}{1-c}\vec{v}\vec{v}^T=I+\frac{-na}{1-(-na)}\frac{1}{\sqrt{n}}\begin{pmatrix}1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}\frac{1}{\sqrt{n}}\begin{pmatrix}1 & 1 & \ldots & 1 \end{pmatrix} \\ & =I-\frac{a}{1+na}\begin{pmatrix}1 & 1 & 1 & \ldots & 1\\ 1 & 1 & 1 & \ldots & 1 \\ \vdots & \vdots & \vdots & \ldots & \vdots \ \\ 1 & 1 & 1 & \ldots & 1\end{pmatrix}\end{align*} We could't simplify the form of that matrix, could we? $\endgroup$
    – Mary Star
    Mar 27, 2020 at 15:28
  • $\begingroup$ Not much. You could reuse the matrix $A$ to get $A^{-1} = I-\frac{1}{1+na}(A-I).$ Or you could introduce the vector $\vec{w}=(1\;\ldots\;1)^T$ and write $A^{-1} = I-\frac{a}{1+na}\vec{w}\vec{w}^T,$ if you consider this a simplification. $\endgroup$ Mar 27, 2020 at 15:45

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