Problem Sheet - Linear Algebra MT22, V
Flashcards
\[\left(\begin{matrix} 1 & 0 & \frac{1}{3} & 1 \\ 0 & 1 & \frac{-2}{3} & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right)\]
How would you describe where you put $-1$s for the Minus-1 trick?
In new rows so that there’s always a $1$ or $-1$ on the diagonal.
\[\left(\begin{matrix} 1 & 0 & \frac{1}{3} & 1 \\ 0 & 1 & \frac{-2}{3} & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right)\]
How would you rewrite this matrix for the Minus-1 trick?
\[\left(\begin{matrix} 1 & 0 & \frac{1}{3} & 1 \\ 0 & 1 & \frac{-2}{3} & 0 \\ 0 & 0 & 0 & 0 \end{matrix}\right) \to \left(\begin{matrix} 1 & 0 & \frac{1}{3} & 1 \\ 0 & 1 & \frac{-2}{3} & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right)\]
Here the minus-1 trick has been applied. What are the solutions to the original matrix being $0$?
How can you find the kernel of a linear transformation represented by a matrix $A$?
Write in reduced row-echelon form and use the minus-1 trick to read off solutions.
How can you find the image of a linear transformation represented by a matrix $A$?
Write in reduced column-echelon form and read off the pivot columns.
What does the linear transformation $(S + T)(v)$ mean where $S, T : V \to W$?
What does the linear transformation $(\lambda S)(v)$ mean where $S, T : V \to W$?
@Define what it means for a linear transformation $T$ to be idempotent.
How, in practice, could you get a matrix in reduced column-echelon form?
Take the transpose of the matrix, get in RREF, then take the transpose again.
In what situation might you want to use the reduced column-echelon form?
When determining the image of a matrix.