# Linear algebra: Check the statement

1. Dec 6, 2015

### gruba

1. The problem statement, all variables and given/known data
Check the statement is true or false:
Let $\mathcal{A} : \mathbb{R^3}\rightarrow \mathbb{R^4}$ be a linear operator.
If the minimum rank of $\mathcal{A}$ is $2$, than the maximum defect is $1$.

2. Relevant equations
-Linear transformations

3. The attempt at a solution
Assume that $\mathcal{A}$ is a matrix of order $3$. If the maximum rank of a matrix is two, then number of defects (linearly dependent vectors) is $1$.
Thus, the statement is true.

Is this correct?

2. Dec 6, 2015

### Staff: Mentor

Since $A :\mathbb{R}^3 \to \mathbb{R}^4$, you can say something more about the matrix of this transformation. I.e., how many rows and how many columns.
I've never seen this terminology -- number of defects -- before.
What do you mean by "linearly dependent vectors"? Are you talking about column vectors in the matrix or row vectors in the matrix. Please elaborate on what you mean by "linearly dependent vectors".

3. Dec 6, 2015

### gruba

Defect (of a matrix) is linearly dependent column vector.
In terms of a linear transformation (operator), rank is defined as a dimension of an image of that operator.
I am not sure what is the definition of defect in terms of linear operators, that is why I made assumption of matrix.

Again, it depends how would you reduce the matrix in echelon form (row or column).

4. Dec 6, 2015

### Staff: Mentor

I always do row-echelon form or reduced row-echelon form (RREF).

You haven't said what the dimensions of the matrix of A are...

5. Dec 6, 2015

### gruba

If you do RREF, then you would count dependent column vectors (defects).
I think that dimensions of a matrix of linear operator $\mathcal{A}$ is $3\times 3$.

6. Dec 6, 2015

### Staff: Mentor

No. A is a map from $\mathbb{R}^3$ to $\mathbb{R}^4$, so the matrix for A can't possibly be 3 x 3.