# Number of linearly independent eigenvectors

Can we determine if the following matrix has 3 linearly independent eigenvectors without any calculations? I know that it can be determined by calculating the eigenvalues, but I was wondering if something like that can be concluded just by observing this matrix, since it's a true/false question. Thanks

$$\begin{bmatrix} 1 && 3 && 3 \\ -3 && -5 && -3 \\ 3 && 3 && 1 \end{bmatrix}$$

• Note that the columns all add to the same number, that is, $1$. That tells you immediately that $1$ is an eigenvalue. This is really a follow up to @egreg answer. – B. Goddard May 20 '17 at 14:46
• @B.Goddard okay, so whenever I have a marix whose columns all add to the same number, I can automatically conclude that one eigenvalue is 1? – ivana14 May 20 '17 at 14:47
• Not "1", but whatever they add to. Because when you subtract $\lambda I$, you're subtracting $\lambda$ from the sum of each column, making the column sums of $A-\lambda I$ equal $0$, which means $A-\lambda I$ is singular. – B. Goddard May 20 '17 at 15:28

If you try $A+2I$, you get $$\begin{bmatrix} 3 & 3 & 3 \\ -3 & -3 & -3 \\ 3 & 3 & 3 \end{bmatrix}$$ which has rank $1$, so $-2$ is an eigenvalue with geometric multiplicity $2$.
• @ivana14 The first thing to look for is row sums and column sums. So immediately you get one eigenvalue is $1$. A simple look shows also that an eigenvalue is $-2$. – egreg May 20 '17 at 15:00
• so whenever a matrix has a constant row sum and column sum, I can say that one eigenvalue is $1$? or either row or column sum can be constant? How did you get -2 just by looking at the matrix? Sorry, I am just really confused. thanks – ivana14 May 20 '17 at 15:07
• @ivana14 No: if the columns have constant sum $s$, then $s$ is an eigenvalue; similarly if the rows have constant sum. For $-2$, I observed that adding $2$ to the diagonal gives a rank one matrix. – egreg May 20 '17 at 15:29
• @ivana14 Not only the matrix $A+2I$ has zero determinant: it has rank $1$, so its null space has dimension $3-1=2$. – egreg May 20 '17 at 16:14