# Eigenvalues and eigenvectors computation

Let $T$ be the $n \times n$ matrix with every entry equal to $1$. I computed the E-values and E-vectors as follows and was wondering if it is correct:

Since every component of $Tv$ equals $v_1 + ... +v_n$ it is evident that $n$ is an eigenvalue and $(1)=(1,1,1,1....)$ is a corresponding eigenvector.

Next I observed that every column is a multiple of the first. Hence the null space of $T$ has dimension $n-1$. As far as I understand the null space is the eigenspace corresponding to the eigenvalue $0$ (if it is non-trivial). Therefore to find eigenvectors corresponding to $0$ it is enough to find $n-1$ linearly independent vectors. It is evident that the vectors $v_i$ of the form $1$ at the first component and $-1$ at $i$ for $i>1$ satisfy this requirement.

Thank you for checking my solution.

• What od you mean by $(1)$ as vector? Surely you are not assuming $n=1$? – Marc van Leeuwen Aug 14 '13 at 12:09
• @newb The bulked up appearance of text you wrote is extremely demotivating to read. I suggest adding some line breaks. I don't know about other people, but guessing by the lack of answers to a question as simple as this, I suppose they feel the same as me. – Git Gud Aug 14 '13 at 12:20
• Hrm, I didn't have any trouble at all, and though there is one large paragraph I thought it was very direct. It is good advice though, @newb to be sensitive about using linebreaks to make text readable. – rschwieb Aug 14 '13 at 12:26
• @DonAntonio I tried to correct it. – newb Aug 14 '13 at 12:35
• Hey @GitGud, what's the hyper-sensitivity? If you open a random newspaper or novel you'll find plenty of paragraphs with ten or more lines. Are you suggesting you wouldn't even consider reading them? (I even write comments to MSE questions that come close to that, within the limit of the allowed number of keystrokes, but I have the excuse that it is impossible break comments into separate paragraphs.) – Marc van Leeuwen Aug 14 '13 at 12:40

Sure, that is a fine way to do it. You correctly noticed that $T(1,\dots,1)=n(1,\dots,1)$ would yield one eigenvalue $n$, and then secondly noted that the rank of the matrix is obviously $1$, so that the rest of the eigenvalues are $0$. The candidate eigenvectors for eigenvalue $0$ are also straightforwardly chosen. Very efficient :)