Exercise VS.M14

Let \(\mathbf{V}\) be the set \(\mathrm{M}_{\mathrm{22}}\) with the usual addition, but with scalar multiplication defined by
\(\alpha\)A = \(\mathrm{o}_{\mathrm{2,2}}\), for all 2×2 matrices A and scalars \(\alpha\). Determine whether or not \(\mathbf{V}\)
is a vector space with these operations.

Answer

Since only scalar multiplication was changed, we can say that the addition properties for \(\mathbf{V}\) in a vector space were not affected.

Checking the vector space properties of \(\mathbf{V}\) for the new scalar multiplication operation -

1.SC Scalar Closure

If \(\alpha\in\mathbb{C}\) and \(\mathbf{u}\in\mathbf{v}\), then \(\alpha\mathbf{u}\in\mathbf{V}\). Since \(\alpha\)A = \(\mathrm{o}_{\mathrm{2,2}}\), for all 2×2 matrices A and scalars \(\alpha\),
We can say that \(\mathrm{o}_{\mathrm{2,2}}\in\mathbf{V}\) by the axiom Z Zero Vector.
Therefore, the Scalar Closure Property of Vector Space holds true.

2. SMA Scalar Multiplication Associativity

If \(\alpha, \beta\in\mathbb{C}\) and \(\mathbf{u}\in\mathbf{v}\), then \(\alpha(\beta\mathbf{u})=(\alpha\beta)\mathbf{u}\).
Since \(\alpha(\beta\mathbf{u})=\mathrm{o}_{\mathrm{2,2}}\) and \((\alpha\beta)\mathbf{u}=\mathrm{o}_{\mathrm{2,2}}\),
then we can say that the Scalar Multiplication Associativity Property for the Vector Space holds true.

3. DVA Distributivity across Vector Addition

If \(\alpha\in\mathbb{C}\) and \(\mathbf{u},\mathbf{v}=\mathbf{V}\), then \(\alpha(\mathbf{u}+\mathbf{v})=\alpha\mathbf{u}+\alpha\mathbf{v}\).
Since \(\alpha(\mathbf{u}+\mathbf{v})=\mathrm{o}_{\mathrm{2,2}}\) and \(\alpha\mathbf{u}+\alpha\mathbf{v}= \mathrm{o}_{\mathrm{2,2}}\), then we can say that the Distributivity across Vector Addition Property for the Vecotry Space holds true.

4. O One
If \(\mathbf{u}\in\mathbf{v}\), then 1\(\mathbf{u}=\mathbf{u}\).
Since 1\(\mathbf{u}=\mathrm{o}_{\mathrm{2,2}}\) but \(\mathbf{u}\ne\mathrm{o}_{\mathrm{2,2}}\) for all cases except the Zero vector,
then we can say that Property O One for the Vector Space does NOT hold true.

We can then conclude that the \(\mathbf{V}\) defined by the set \(\mathrm{M}_{\mathrm{22}}\) is NOT in the vector space.

.