# For which $a$ does it hold that $\det(A)=0$?

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 $$A=\begin{pmatrix}1+a & a & \ldots & a \\ a & 1+a & \ldots & a \\ \vdots & \vdots & \vdots & \vdots \\ a & a & \ldots & 1+a\end{pmatrix}$$.



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?



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?



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?

• Mar 27, 2020 at 10:55

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.

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...

• 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? Mar 27, 2020 at 11:23

$$\frac{1}{a}A=\frac{1}{a}I-vv^T$$ 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 $$1/a=1\implies a=1$$.

• Why are the only possible eigenvalues $1$ and $0$ ? Mar 27, 2020 at 13:59
• Let $v=(x_1,x_2,...,x_n)^T$ then characteristic equation of $vv^T$ is $t^n-||v||^2t^{n-1}=0$ where $||v||=\sqrt{x_1^2+x_2^2+....+x_n^2}$. Mar 27, 2020 at 14:07

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 $$c= -na.$$

• 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? Mar 27, 2020 at 15:28
• 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. Mar 27, 2020 at 15:45