rove that the image and the kernel of a linear mapping α : V → W are subspaces of V or W . Inyour answer, you should state clearly whether the image or the kernel is a subspace of V or W . Inyour proof, you may use the definitions of the image and the kernel and the standard properties of asubspace. You may refer to the fact that each linear mapping maps the zero vector to the zero vector(which you proved in question 7).
Question
rove that the image and the kernel of a linear mapping α : V → W are subspaces of V or W . Inyour answer, you should state clearly whether the image or the kernel is a subspace of V or W . Inyour proof, you may use the definitions of the image and the kernel and the standard properties of asubspace. You may refer to the fact that each linear mapping maps the zero vector to the zero vector(which you proved in question 7).
Solution
To prove that the image and the kernel of a linear mapping α : V → W are subspaces of V or W, we need to understand the definitions of the image and the kernel.
The image of a linear mapping, also known as the range, is the set of all vectors that can be obtained by mapping the vectors from the domain (V) to the codomain (W). The image is a subset of W.
The kernel of a linear mapping is the set of all vectors in the domain (V) that map to the zero vector in W. The kernel is a subset of V.
To prove that these are subspaces, we need to show that they satisfy the three properties of a subspace:
- The zero vector is in the subspace.
- The subspace is closed under vector addition.
- The subspace is closed under scalar multiplication.
For the image:
- Since every linear mapping maps the zero vector to the zero vector, the zero vector of W is in the image.
- If w1 and w2 are in the image, then there exist v1 and v2 in V such that α(v1) = w1 and α(v2) = w2. Then α(v1 + v2) = α(v1) + α(v2) = w1 + w2, so the image is closed under addition.
- If w is in the image and c is a scalar, then there exists a v in V such that α(v) = w. Then α(cv) = cα(v) = cw, so the image is closed under scalar multiplication.
For the kernel:
- Since every linear mapping maps the zero vector to the zero vector, the zero vector of V is in the kernel.
- If v1 and v2 are in the kernel, then α(v1) = 0 and α(v2) = 0. Then α(v1 + v2) = α(v1) + α(v2) = 0 + 0 = 0, so the kernel is closed under addition.
- If v is in the kernel and c is a scalar, then α(v) = 0. Then α(cv) = cα(v) = c*0 = 0, so the kernel is closed under scalar multiplication.
Therefore, the image is a subspace of W and the kernel is a subspace of V.
Similar Questions
We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU.A part of the proof went as follows. Assuming dim Kerα = m, there is a basis of Kerα consisting ofm vectors; denote them by u1, . . . , um. Since dim Imα = n, there is a basis of Imα consisting of nvectors; denote them by v1, . . . , vn. Since v1 is in Imα, there is w1 ∈ U such that α(w1) = v1, andsimilarly for all other vectors in the set v1, . . . , vn; in this way, we define n vectors w1, . . . , wn.Now consider a vector u ∈ U . Since α(u) is in Imα, we have α(u) = d1v1 + · · · + dnvn for somenumbers d1, . . . , dn. Consider a vector u′ ∈ U defined as u′ = d1w1 + · · · + dnwn.Prove that u − u′ ∈ Kerα.
We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU
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