Range of a linear transformation to power n

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To prove that the range of a square matrix A raised to the power of n+1 is a subspace of the range of A raised to the power of n, one can use the relationship A^{n+1}x = A^n(Ax). This shows that any vector y in the range of A^{n+1} can be expressed as A^n multiplied by some vector Ax, indicating that y is also in the range of A^n. The discussion also touches on proving that the kernel of A^n is a subspace of the kernel of A^{n+1}. The participants are focused on establishing the properties of ranges and kernels in the context of linear transformations. The conversation concludes with a clear understanding of the implications of the transformation relationships.
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Homework Statement



How to prove that: the range of a square matrix A (linear transformation) to the power of n+1 is a subspace of the Range of A to the power n, for all n >= 1?

i.e. Range (A^(n+1)) is a subspace of Range (A^n)

Homework Equations





The Attempt at a Solution



I can prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)). Not sure if this is the basis of the prove.
 
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Thank you jbunniii. Using the hint you gave me I could prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)), because A^n(A(0)) = 0. but what about range? I think if i could prove the range of A is always a subspace of domain of A, then i am done. How to prove this? Thanks.
 
If y is in the range of A^{n+1}, then by definition,

y = A^{n+1}x

for some x.

But

y = A^n(Ax)

So what does that imply?
 
Got it. Thanks so much.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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