(True and False Confirmation)

Question

1) True because This is true by definition of an invertible matrix/any arbitrary matrix

2) True because a linear map T̃ :V→V And the image of T̃ T~ is by definition the image which is all of V, T~ is surjective. Then the kernel of T̃ T~ is {0}, and by definition this kernel is the intersection of the kernel of T with its image V.

3)True because A is diagonalizable if and only if A has n linearly independent eigenvectors.

4)True because kernel is just the set of all eigenvectors of (or ) associated with theeigenvalue plus the zero vector

5)True because A^−1=(Q^TDQ)^−1=Q^−1D^−1(Q^T)^−1=Q^TD^−1Q.

6)False reflection y-axis with rotation cant form an upper tirangular

7)False

8) False three dimensional non-linear subspaces are not contained

Answer 1

1. T.
2. $\times$ F. We can only know that $\dim (\ker (\mathcal T)) + \dim(\mathrm {im} (\mathcal T)) = \dim (V)$, but it doesn't mean the intersection is $\{0\}$. Counterexample: as @Alon Amit stated.
3. $\times$ F. Try identity matrix.
4. T. Not your explanation. We have $\mathcal Av = c v$. If $c = 0$, then $v \in \ker (\mathcal A)$. If $c \neq 0$, then $cv \in \mathrm{im}(\mathcal A)$, and so is $v = cv /c$.
5. T.

6. F. But this is not a $y$-reflection. This is a rotation. In $\mathrm M_2(\mathbb R)$, the rotation matrices cannot be diagonalized.

7. $\times$ T. By row operation, $\boldsymbol{AB} \sim \boldsymbol {IB}= {\boldsymbol B}$, which is a process of left multiplication by a series of elementary matrices. Therefore $\boldsymbol {AB}, \boldsymbol B$ has the same reduced row echelon form.

8. $\times$ T. Here the subspaces we are talking about are linear subspaces. This is true, but not so easy to explain. There is a proposition: if the base is infinite field, then any linear space cannot be covered by a finite collection of proper subspaces. Geometrically, you can find infinitely many straight lines passing through the origin in $\mathbb R^2$ where each line is a subspace of $\Bbb R^2$ This claim is an analogy.

Answer 2

Many of your answers are incorrect, not so much for linear algebraic reasons but for reasons of logic, or language. For example, the intersection of kernel and image of a linear operator need not be $\{0\}$, and your explanation makes – I’m sorry – no sense at all.

(For example, the linear operator $T:\mathbb{R}^2\to \mathbb{R}^2$ given by $T((a,b))=(0,a)$ has a one-dimensional Image and a one-dimensional kernel which are the same.)

The last question asserts that any four-dimensional space has infinitely many three-dimensional spaces. This is true over infinite ground fields, but not true for finite dimensional spaces over finite fields. We can’t tell if your course or textbook consider finite fields as possible ground fields.

But the bigger issue is that your answer – “non-linear spaces aren’t contained” – is very far off the mark. The question wasn’t whether a certain class of spaces are all contained in the space, and it’s unclear what spaces that aren’t contained in our space have to do with the question.