Undergrad How can I convince myself that I can find the inverse of this matrix?

[Mark44]

The first is a row vector. The second is a column vector, which is the transpose of the row vector. IOW, the column vector is $x^T$, using the usual notation.

I thought it was the other way around.

Not sure what you're considering to be the other way around. My reply to @Hall, which is quoted above, was a response to his asking what is the difference between $[x_1, x_2, \dots, x_n]$ and $\begin{bmatrix}x_1 \\ x_2 \\ \dots \\ x_n \end{bmatrix}$.
Rows are horizontal and columns (like the columns of a building are vertical, so the first vector above is a row vector, and the second vector is a column vector.

If your confusion is with the notation $x^T$, a transpose can be either a row vector or a column vector, depending on how $x$ is originally defined.

 

[swampwiz]

Not sure what you're considering to be the other way around. My reply to @Hall, which is quoted above, was a response to his asking what is the difference between $[x_1, x_2, \dots, x_n]$ and $\begin{bmatrix}x_1 \\ x_2 \\ \dots \\ x_n \end{bmatrix}$.
Rows are horizontal and columns (like the columns of a building are vertical, so the first vector above is a row vector, and the second vector is a column vector.

If your confusion is with the notation $x^T$, a transpose can be either a row vector or a column vector, depending on how $x$ is originally defined.

What I was saying was the I thought the nominal form of a vector is as a column, not as a row. Certainly if written as A x, x is a column vector.

 

[Mark44]

What I was saying was the I thought the nominal form of a vector is as a column, not as a row. Certainly if written as A x, x is a column vector.

I don't think there is a nominal form of a vector. However, in the context of the expression Ax, with A being a matrix, x would have to be a column vector.

 

[WWGD]

More generally, given an $n \times n$ matrix, the vector must be $n \times 1$ for the product to be defined. So, yes, a column vector.

 

[swampwiz]

What about the eigenproblem equation? A x = 0, but the eigenvectors are solutions of x that are NOT 0; of course, this is possible because the determinant of A must be 0 for this to work.

The eigenproblem EQ is [ A ]{ x } = λ { x }, which leads to [ [ A ] - λ [ I ] ] { x } = { 0 }, and only works for the case of Δ( [ [ A ] - λ [ I ] ] ) = 0. Eigenvectors correspond to the nullspace of a matrix.