# Relation between eigenvalues and rows or columns of a matrix

I am wondering about the relation between the eigenvalues of a matrix $A$ and it rows or columns. I mean can one say that $\lambda_1$ is related to $r_1$ (first row of $A$) or $c_1$ (first column of $A$)? I know you may ask then what is $\lambda_1$? But the question is, what is the relation between a specific eigenvalue of a matrix and a column or a row of the matrix?

If there is a relation between rows (or columns) and eigenvalues. Say that: $\lambda_1$ corresponds to $r_1$. If i take the transpose of A, since eigenvalues still are same, can I still say $\lambda_1$ corresponds to $r_1'$ (first row of $A'$).

This might be a little strange question, but i appreciate for any ideas.

-

Let's break this down (I'm going for more of an intuitive answer here):

When you multiply a matrix $M$ and a vector $v_i$, you take each row of the matrix and do an inner product with the vector to get the elements of the resulting vector $v_o$ -- so row 1 of $M$ times $v_i$ gets you the first element in $v_o$, row 2 of $M$ gets you the second element of $v_o$, etc.

If $v_i$ is an eigenvector of $M$, then multiplying them will give you a $v_o$ that is effectively $v_i$ times a constant, where the constant is the eigenvalue corresponding to the eigenvector.

Since the whole output vector ($v_o$) is a scalar multiple of the input vector ($v_i$), the eigenvalues cannot be directly related to specific rows of $M$; each element of $v_o$ only "interacted" with a single row of $M$ -- a different row for each element -- during the multiplication, but every element of $v_o$ was scaled by the same (eigen)value from the original elements in $v_i$.

As for the columns, every column of $M$ affects every element of the output vector, so how do we separate out the distinct eigenvalues?

The whole of $M$ can be seen as a transformation. For certain vectors, the transformation only ends up scaling that vector -- this is just a result of applying a particular transformation to a particular vector.

-

Well, spectral life of matrices is not that straightforward and simple, unfortunately.

For a given (square) matrix $M$, one can find an equivalent matrix $M'=B^{-1}MB$ for some invertible $B$ matrix, which is in Jordan normal form, that is, almost diagonal. And its diagonal elements (say, columns as you wish) are just the eigenvalues.

And, practically this $B$ matrix contains among its columns one eigenvector for each eigenvalue (counted with multiplicity)..

Considering columns or rows corresponds to the question, from which direction we want the matrix to act on vectors. Meant, $Mv$ can be written for column vector $v$, and $uM$ for row vector $u$...

-