A3 is a 3 x 3 matrix

Find the determinant of each of those matrices and eventually you'll notice the relationship in a characteristic equation:

[tex]A = \left(\begin{array}{abcdef}
1-\lambda & 1 & 1\\
1 & 1-\lambda & 1\\
1 & 1 & 1-\lambda
\end{array}\right)[/tex]

[tex]det(A) = -\lambda^3+3\lambda^2=0 [/tex]

Similarly for [tex]A_{2x2}, det(A)=\lambda^2-2\lambda=0[/tex]
[tex]A_{4x4}, det(A)=\lambda^4-4\lambda^3=0[/tex]
[tex]A_{5x5}, det(A)=-\lambda^5+5\lambda^4=0[/tex]


The [tex]\lambda[/tex] is the eigenvalue. You can find your associated eigenvector by solving for this equation:

[tex]Av=\lambda v[/tex]

Where A is the matrix with all entries 1, v is the eigenvector, and [tex]\lambda[/tex] is the eigenvalue.

For example:

[tex]A_{2x2}:[/tex]

[tex]A = \left(\begin{array}{abcdef}
1-\lambda & 1\\
1 & 1-\lambda\\
\end{array}\right)[/tex]

[tex]det(A) = (1-\lambda)(1-\lambda)-1=0[/tex]

Solving for [tex]\lambda[/tex] we get: [tex]\lambda=0, 2[/tex]

Now find associated eigenvectors for each eigenvalue:
[tex]Since Av=\lambda v[/tex]
[tex]Av-\lambda v=0[/tex]
[tex](A-\lambda)v=0[/tex]
v is your eigenvector

So for [tex]\lambda=0:[/tex]

[tex]A = \left(\begin{array}{abcdef}
1-0 & 1 \\
1 & 1-0 \\
\end{array}\right)[/tex]

[tex]A = \left(\begin{array}{abcdef}
1 & 1 \\
1 & 1 \\
\end{array}\right)[/tex]

Your eigenvector is of the form [tex]v = \left(\begin{array}{abcdef}
v_1 \\
v_2 \\
\end{array}\right)[/tex]

Multiplying out with vector v and equationg to 0 you get:
[tex] 1v_1 + 1v_2 = 0[/tex]
[tex] 1v_1 + 1v_2 = 0[/tex]

Therefore in this particular case [tex]1v_1 = -1v_2[/tex] you can pick any number for [tex]v_2[/tex], i'd go with [tex]v_2=1[/tex]. So your eigenvector for [tex]\lambda=0[/tex] is:

[tex]v = \left(\begin{array}{abcdef}
-1 \\
1 \\
\end{array}\right)[/tex]

You can check your eigenvalues and eigenvector simply by multiplying them out, since [tex]Av=\lambda v[/tex]

Check:

[tex] \left(\begin{array}{abcdef}
1 & 1 \\
1 & 1 \\
\end{array}\right)[/tex][tex]\left(\begin{array}{abcdef}
-1 \\
1 \\
\end{array}\right)[/tex][tex]=0*\left(\begin{array}{abcdef}
-1\\
1 \\
\end{array}\right)[/tex]

However to answer your question for [tex]A_{nxn}[/tex] matrices..

since det(A)=0 for all such matrices and trace(A)=n you may notice that for every [tex]A_{nxn} [/tex] matrix you get n eigenvalues, but only one of them is an real number eigenvalue, [tex]\lambda=n[/tex]. In other words, for [tex]A_{2x2}, \lambda=2,0; A_{3x3}, \lambda=3,0,0; A_{4x4}, \lambda=4,0,0,0[/tex] --- you get the idea.

 

with no calculation at all, just thinking abut the meaning of eigenvalues, one immediately sees that there is an n-1 dimensional kernel, spanned by n-1 independent eigenvectors, namely e1-e2, e1-e3,....,e1-en.

then since every one of these standard vectors goes to (1,...,1), their sum which equals this vector, goes to (n,...,n), so e1+...+en is another eigenvector.

the moral here is to think about the meaning of things before calculating. this approach led me to see the answer almost instantly, even before getting as far as your EDIT, much less cronxeh's post.

 

Not to be a pain in the butt, but you forgot the idenity matrix (A-[tex]\lambda[/tex]I)