How can I prove the rank of a matrix with a specific pattern of entries?

[A.Magnus]

I would love to get help on this problem: Suppose that $M$ is a square $k \times k$ matrix with entries of 1's in the main diagonal and entries of $\frac{1}{k}$ for all others. Show that the rank of $M$ is $k$.

I think I should go about by contradiction, that is, by assuming that the column vectors are not linearly independent. Since there are $k$ number of column vectors, then hopefully I can show that the rank of $M$ is indeed $k$. Unfortunately I don't know how to put these ideas down in writing; your time and gracious helps are therefore very much appreciated. Thank you - MA

 

[I like Serena]

I would love to get help on this problem: Suppose that $M$ is a square $k \times k$ matrix with entries of 1's in the main diagonal and entries of $\frac{1}{k}$ for all others. Show that the rank of $M$ is $k$.

I think I should go about by contradiction, that is, by assuming that the column vectors are not linearly independent. Since there are $k$ number of column vectors, then hopefully I can show that the rank of $M$ is indeed $k$. Unfortunately I don't know how to put these ideas down in writing; your time and gracious helps are therefore very much appreciated. Thank you.

Hi MaryAnn! Welcome to MHB! ;)

Suppose we add all column vectors together and divide by $1+\frac{k-1}k$, then we get:
$$\begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}$$
Now subtract $k$ times the first column vector:
$$\begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}
- k\begin{bmatrix}1\\ 1/k\\ \vdots \\ 1/k\end{bmatrix}
=\begin{bmatrix}1-k\\ 0\\ \vdots \\ 0\end{bmatrix}
$$
Hey! That's a multiple of a standard unit vector!
Moreover, we can get each unit vector from a linear combination of the column vectors.
Therefore the column vectors are independent.

 

[A.Magnus]

Hi MaryAnn! Welcome to MHB! ;)

Suppose we add all column vectors together and divide by $1+\frac{k-1}k$, then we get:
$$\begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}$$
Now subtract $k$ times the first column vector:
$$\begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}
- k\begin{bmatrix}1\\ 1/k\\ \vdots \\ 1/k\end{bmatrix}
=\begin{bmatrix}1-k\\ 0\\ \vdots \\ 0\end{bmatrix}
$$
Hey! That's a multiple of a standard unit vector!
Moreover, we can get each unit vector from a linear combination of the column vectors.
Therefore the column vectors are independent.

Thank you for your gracious help! This is more than just a genius' solution. - MA