Linear algebra: Check the statement

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Homework Help Overview

The discussion revolves around a statement regarding a linear operator \(\mathcal{A} : \mathbb{R^3}\rightarrow \mathbb{R^4}\) and its rank and defect. Participants are examining the implications of the minimum rank being 2 and whether this leads to a maximum defect of 1.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to clarify the definitions of rank and defect in the context of linear transformations and matrices. Questions arise about the terminology used, particularly regarding "linearly dependent vectors" and how it relates to the dimensions of the matrix.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the terms involved. There is a lack of consensus on the definitions and implications of the terms "defect" and "linearly dependent vectors." Some guidance on the relationship between rank and defect has been offered, but clarity on the dimensions of the matrix remains unresolved.

Contextual Notes

Participants note the confusion surrounding the dimensions of the matrix associated with the linear operator, with some asserting it to be \(3 \times 3\) while others argue it must reflect the mapping from \(\mathbb{R}^3\) to \(\mathbb{R}^4\).

gruba
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Homework Statement


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.

Homework Equations


-Linear transformations

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?
 
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gruba said:

Homework Statement


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.

Homework Equations


-Linear transformations

The Attempt at a Solution


Assume that \mathcal{A} is a matrix of order 3.
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.
gruba said:
If the maximum rank of a matrix is two, then number of defects (linearly dependent vectors) is 1.
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".
gruba said:
Thus, the statement is true.

Is this correct?
 
Mark44 said:
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".

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).
 
gruba said:
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).
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...
 
Mark44 said:
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...

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.
 
gruba said:
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.
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.
 

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