Click the arrow to see the answer.
Let be a basis for Then there is a basis for and which are respectively It follows that you must have and so you must have
Click the arrow to see the answer.
Let be a basis for Then there is a basis for and which are respectively It follows that you must have and so you must have
These problems came from Chapter 6 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 358–359, and p. 370.
2024-09-06 23:47:04