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Let A be a square matrix which is diagonalizable over field $\mathbb{F}$ and the sum of the entries of any column is the same number $a\in\mathbb{F}$. Show that $a$ is eigenvalue of the matrix A.

My try:

Let A be the arbitrary matrix $\displaystyle \left(\begin{matrix} a_{11}&a_{12}&\cdots&a_{1(n-1)}&a_{1n} \\ a_{21}&a_{22}&\cdots&a_{2(n-1)}&a_{2n} \\ \vdots&\vdots & \ddots&\vdots&\vdots \\a_{(n-1)1}&a_{(n-1)2}&\cdots&a_{(n-1)(n-1)}&a_{(n-1)n} \\ a_{n1}&a_{n2}&\cdots&a_{n(n-1)}&a_{nn}\end{matrix}\right)$.

Now $\displaystyle p_A(x)=\det(xI-A)=\det\left(\begin{matrix} x-a_{11}&-a_{12}&\cdots&-a_{1(n-1)}&-a_{1n} \\ -a_{21}&x-a_{22}&\cdots&-a_{2(n-1)}&-a_{2n} \\ \vdots&\vdots & \ddots&\vdots&\vdots \\-a_{(n-1)1}&-a_{(n-1)2}&\cdots&x-a_{(n-1)(n-1)}&-a_{(n-1)n} \\ -a_{n1}&-a_{n2}&\cdots&-a_{n(n-1)}&-a_{nn}\end{matrix}\right)$.

Multiply row and add it to other row doesn't change the value of the determinant, hance we can add any row $2\le j\le n$ to the first row and get $$p_A(x)=\det\left(\begin{matrix} x-\sum_{i=1}^na_{i1}&x-\sum_{i=1}^na_{i2}&\cdots&x-\sum_{i=1}^na_{i(n-1)}&x-\sum_{i=1}^na_{in} \\ -a_{21}&x-a_{22}&\cdots&-a_{2(n-1)}&-a_{2n} \\ \vdots&\vdots & \ddots&\vdots&\vdots \\-a_{(n-1)1}&-a_{(n-1)2}&\cdots&x-a_{(n-1)(n-1)}&-a_{(n-1)n} \\ -a_{n1}&-a_{n2}&\cdots&-a_{n(n-1)}&-a_{nn}\end{matrix}\right)$$ But we know that $\displaystyle \forall 1 \le k \le n: \ \sum_{i=1}^n a_{ik}=a$, hence $$p_A(x)=\det\left(\begin{matrix} x-a&x-a&\cdots&x-a&x-a \\ -a_{21}&x-a_{22}&\cdots&-a_{2(n-1)}&-a_{2n} \\ \vdots&\vdots & \ddots&\vdots&\vdots \\-a_{(n-1)1}&-a_{(n-1)2}&\cdots&x-a_{(n-1)(n-1)}&-a_{(n-1)n} \\ -a_{n1}&-a_{n2}&\cdots&-a_{n(n-1)}&-a_{nn}\end{matrix}\right)= \\ =(x-a)\cdot\det\left(\begin{matrix} 1&1&\cdots&1&1 \\ -a_{21}&x-a_{22}&\cdots&-a_{2(n-1)}&-a_{2n} \\ \vdots&\vdots & \ddots&\vdots&\vdots \\-a_{(n-1)1}&-a_{(n-1)2}&\cdots&x-a_{(n-1)(n-1)}&-a_{(n-1)n} \\ -a_{n1}&-a_{n2}&\cdots&-a_{n(n-1)}&-a_{nn}\end{matrix}\right)$$ thus $a$ is an eigenvalue of the matrix A.

I didn't use the fact that A is diagonalizable and don't know why it is needed.

Any help/hint will be appreciated, thank you!

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3 Answers 3

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$A$ and $A^T$ have the same eigenvalues.

Take $X=\left( \begin{matrix} 1\\ 1\\ ...\\ 1 \end{matrix} \right)$

$A^T X=aX$, $a$ being the sum of each column of $A$, thus the sum of each row of $A^T$.

Then $a$ is eigenvalue of $A^T$, and of $A$.

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  • $\begingroup$ I understand this answer, but I want to know what is wrong with my solution. $\endgroup$
    – Galc127
    Jan 6, 2015 at 11:07
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There is nothing wrong with the proof although you probably use a gun to kill the fly.

Indeed, you don't need $A$ to be diagonalizable. We can construct a matrix $A$ with this property, which is not diagonalizable, easily. Consider $$ A=\left[\begin{array}{rrr}0&-1&-1\\1&1&0\\0&1&2\end{array}\right]. $$ The column sums are all equal to $1$, which is the only eigenvalue of $A$ with the algebraic multiplicity $3$. However, $A$ is not diagonalizable.

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i will avoid the use of the fact that $A$ and $A^T$ have the same eigenvalue therby avoiding any appeal to determinants explicitly. bus we do use the fact the one sided invertiblilty implies the other. that is $AB = I$ iff $BA = I.$

if the sum of every column is $a,$ then $A$ has left eigenvector corresponding to $a.$ that is there is $u = (1,1,\cdots, 1)^T \neq 0$ such that $u^TA = au^T.$ therefore $A - aI$ is not invertible which in turn implies that $a$ is an eigenvalue of $A$ and has a (right)eigenvector $x \neq 0$ such that $Ax = ax$

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  • $\begingroup$ Hmm, but why is a left eigenvalue also a right eigenvalue? You don't need determinant to explain this, of course, but at the end of the day, you still need to explain this left-right thing. +1 anyway. $\endgroup$
    – user1551
    Jan 6, 2015 at 9:09
  • $\begingroup$ @user1551, because $A-aI$ is invertible and one invertibilty implies the other. that is $AB = I$ iff $BA = I$ $\endgroup$
    – abel
    Jan 6, 2015 at 9:11
  • $\begingroup$ I understand this answer, but I want to know what is wrong with my solution. $\endgroup$
    – Galc127
    Jan 6, 2015 at 11:06
  • $\begingroup$ @abel But $A-aI$ is not invertible. You say that there is a nonzero $u$ (a constant vector) s.t. $u^T(A-aI)=0$ which is equivalent to the fact that there is a nonzero $v$ s.t. $(A-aI)v=0$, that is, $A^T-aI$ is not invertible iff $A-aI$ is not invertible, which turns out to be exactly the same thing as saying that $A$ and $A^T$ have the same eigenvalues. $\endgroup$ Jan 6, 2015 at 12:00
  • $\begingroup$ @AlgebraicPavel, i meant to write $A-aI$ not invertible in the comment. can you claim $A$ and $A^T$ have the same eigenvalues without using $det(A) = det( A^T)?$ by the way, i like most of your answers, precise and brief, to matrix questions. $\endgroup$
    – abel
    Jan 6, 2015 at 14:58

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